Statistics Exam 3

A random sample of 1019 adults in a certain large country was asked​ "Do you pretty much think televisions are a necessity or a luxury you could do​ without?" Of the 1019 adults​ surveyed, 527 indicated that televisions are a luxury they could do without. a) Obtain a point estimate for the population proportion of adults in the country who believe that televisions are a luxury they could do without. B)The sample--- asimple random sample the value of --- is --which is --- 10 and the --- ---- less than or equal to 5% of the ---- np^q^= (1019)(.517)(1-.517)=254.456 c) ​Construct and interpret a 95​% confidence interval for the population proportion of adults in the country who believe that televisions are a luxury they could do without. d) Is it possible that a supermajority​ (more than​ 60%) of adults in the country believe that television is a luxury they could do​ without? Is it​ likely? It is possible but not likely that a supermajority of adults in the country believe that television is a luxury they could do without because the 95​% confidence interval does not conatin 0.60. e) Use the results of part​ (c) to construct a 95​% confidence interval for the population proportion of adults in the country who believe that televisions are a necessity

(a) 0.517 p^=527/1019 B)is stated to be---np^(1-p^) ---254.456---greater than or equal to---sample size can be assumed to be--- population size. c)We are 95% confident the proportion of adults in the country who believe that televisions are a luxury they could do without is between .486 and .548. p^=.517 α=1-.95=0.05 Z0.05/2=1.96 E=1.96sr(pq/n) =1.96sr(.517)(.483)/1019=0.031 UB=.517+.031=.548 LB=.517-.031=.486 d) CI=.486-P^-.548 p=.60? Yes possible b/c we are never 1000% confident e) UB=.514 LB=.452 1-oldUB=1-.548=.452 1-oldLB=1-.486=.514

A simple random sample of size n=450 individuals who are currently employed is asked if they work at home at least once per week. Of the 450 employed individuals​ surveyed, 42 responded that they did work at home at least once per week. Construct a​ 99% confidence interval for the population proportion of employed individuals who work at home at least once per week.

0.058 to 0.128 42/450=.093 (1-.99)/2=.005 invnorm(1-.005)=2.576 0.093-(2.576xsr[(.093(1-.093)/450]=0.058 0.093+(2.576 x sr[(.093(1-.093)/450]=0.128

A survey was conducted that asked 1005 people how many books they had read in the past year. Results indicated that x=14.3 books and s=16.6 books. Construct a 95​% confidence interval for the mean number of books people read. Interpret the interval.

13.27 to 15.33 Tα/2=(1-.95)/2=.025 invT(1-.025, 1004)=1.962 LB:14.3-1.962(16.6/sr1005) ub:14.3+1.962(16.6/sr1005) There is 95​% confidence that the population mean number of books read is between 13.27 and 15.33.

State the conclusion based on the results of the test. According to the​ report, the standard deviation of monthly cell phone bills was ​$49.34 three years ago. A researcher suspects that the standard deviation of monthly cell phone bills is different today. The null hypothesis is not rejected.

There is not sufficient evidence to conclude that the standard deviation of monthly cell phone bills is different from its level three years ago of ​$49.34.

A random sample of n1=125 individuals results in x1=40 successes. An independent sample of n2=150 individuals results in x2=60 successes. Does this represent sufficient evidence to conclude that p1<p2 at the α=0.05 level of​ significance? a) What type of test should be​ used? (b) Determine the null and alternative hypotheses ​(c) Use technology to calculate the​ P-value. z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2 d) Draw a conclusion based on the hypothesis test.

A) A hypothesis test regarding the difference between two population proportions from independent samples. b)H0:p1=p2 H1:p1<p2 c)p=0.085 calculator: stats, test, (6) 2-PropzTest, x1=122​, n1=254​, x2=134​, and n2=317, p1>p2 d)0.085>0.05 There is not sufficient evidence to reject the null hypothesis because the ​P-value>α.

Construct a 95​% confidence interval of the population proportion using the given information. x=125, n=250 A) lower bound B) Upper bound p=125/250=0.5 α=0.05 (zα/2)=0.05/2=z0.025=1.96 E=(zα/2)sr(pq/n) E=1.96sr(.5x.5/250)=.06198

A).438 LB=p-E=.5-.06198 B).562 UP=p+E=.5+.06198=.56198

According to a certain government agency for a large​ country, the proportion of fatal traffic accidents in the country in which the driver had a positive blood alcohol concentration​ (BAC) is 0.32. Suppose a random sample of 114 traffic fatalities in a certain region results in 46 that involved a positive BAC. Does the sample evidence suggest that the region has a higher proportion of traffic fatalities involving a positive BAC than the country at the α=0.1 level of​ significance? a)What are the null and alternative​ hypotheses? b) Find the test​ statistic, z0. c)p value d)Determine the conclusion for this hypothesis test. Choose the correct answer below.

Because npq=24.8>10,the sample size is less than ​5% of the population​ size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied. a)H0:p=.32 vs H1:p>.32 b)1.91 46/114=.404 (.404-.32)/sr((.32x.68)/114) c)0.028<0.1 normalcdf(1.92, 999) .028 d)Since ​P-value<α​, reject the null hypothesis and conclude that there is sufficient evidence that the region has a higher proportion of traffic fatalities involving a positive BAC than the country

Suppose a researcher is testing the hypothesis H0​:p=0.4 vs H1​:p>0.4 and she finds the​ P-value to be 0.23. a)Explain what this means. Would she reject the null​ hypothesis? Why?

If the​ P-value for a particular test statistic is 0.23​, she expects results at least as extreme as the test statistic in about 23 of 100 samples if the null hypothesis is true. b)Since this event is not​ unusual, she will not reject the null hypothesis.

A simple random sample of size n=40 is drawn from a population. The sample mean is found to be x=121.4 and the sample standard deviation is found to be s=13.2. Construct a​ 99% confidence interval for the population mean.

Lb=115.75 ub=127.05 Tα/2=(1-.99)/2=.005 invT(1-.005, 39)=2.708 LB:121.4-2.708(13.2/sr40) UB:121.4+2.708(13.2/sr40)

A simple random sample of size n=19 is drawn from a population that is normally distributed. The sample mean is found to be x=70 and the sample standard deviation is found to be s=19. Construct a 90​% confidence interval about the population mean.

Lb=62.44 ub=77.56 Tα/2=(1-.90)/2=.05 invT(1-.05, 18)=1.734 LB:70-1.734(19/sr19) UB:70+2.08(19/sr19)

A trade magazine routinely checks the​ drive-through service times of​ fast-food restaurants. An 80​% confidence interval that results from examining 515 customers in one​ fast-food chain's​ drive-through has a lower bound of 167.8 seconds and an upper bound of 171.2 seconds. What does this​ mean?

One can be 80​% confident that the mean​ drive-through service time of this​ fast-food chain is between 167.8 seconds and 171.2 seconds.

Explain what​ "statistical significance" means.

Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true.

Explain the difference between statistical significance and practical significance.

Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application.

In a​ poll, 69​% of the people polled answered yes to the question​ "Are you in favor of the death penalty for a person convicted of​ murder?" The margin of error in the poll was 3​%, and the estimate was made with 95​% confidence. At least how many people were​ surveyed?

The minimum number of surveyed people was 914. zα/2=(1-.95)=.05 .05/2=.025=1-.025=.975 invonorm(.975)=1.96 n=p^q^((zα/2)/E)^2 p=.69 q=.31 E=.03 zα/2=1.96 n=(.69x.31)(1.96/.03)^2=914

Construct a confidence interval for p1−p2 at the given level of confidence. x1=32​, n1=248​, x2=36​, n2=300​,99​% confidence

The researchers are ​99% confident the difference between the two population​ proportions, p1−p2​, is between -0.064 and 0.082 calculator: stats, test, (B) 2-Prop2Int x1=32​, n1=248​, x2=36​, n2=300​ clevel=0.99 anser is interval

construct a confidence interval for p1−p2 at the given level of confidence. x1=27​, n1=265​, x2=32​, n2=310​, 90​% confidence z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2

The researchers are 90% confident the difference between the two population​ proportions, p1−p2​, is between -0.043 and 0.04 stats, test, (B) 2-Prop2int Clevel=.90

A researcher wishes to compare personality types of patients with and without narcolepsy. She obtains a random sample of 61 patients of each category who take a personality inventory and determines each patient's personality type.

The sampling is independent b/c an individual selected for one sample does not dictate which individual is to be the second sample. The variable is qualitative b/c it classifies the individual

In a survey of 1012 ​adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not. Of the 1012 ​surveyed, 534 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.

There is 99% confidence that the true proportion of worried adults is between 0.487 and 0.569. 534/1012=0.528 (1-.99)/2=.005 invnorm(1-.005)=2.576 .528-(2.576xsr[(.528(1-.528)/1012]=0.487 528+(2.576 x sr[(.528(1-.528)/1012]=0.569

State the conclusion based on the results of the test. According to the Federal Housing Finance​ Board, the mean price of a​ single-family home two years ago was ​$299,100. A real estate broker believes that because of the recent credit​ crunch, the mean price has increased since then. The null hypothesis is not rejected.

There is not sufficient evidence to conclude that the mean price of a single-family home has increased from its level two years ago of $299100

Suppose the null hypothesis is rejected. State the conclusion based on the results of the test. Three years​ ago, the mean price of a​ single-family home was ​$243,716. A real estate broker believes that the mean price has decreased since then.Which of the following is the correct​ conclusion?

There is sufficient evidence to conclude that the mean price of a​ single-family home has decreased.

Determine whether the following statement is true or false. To construct a confidence interval about the​ mean, the population from which the sample is drawn must be approximately normal.

This statement is False.

Determine if the following statement is true or false. To perform a​ one-way ANOVA, the populations do not need to be normally distributed.

This statement is false; The requirements for​ one-way ANOVA are shown below. 1. There must be k simple random​ samples, one from each of k populations or a randomized experiment with k treatments. 2.The k samples must be independent of each​ other; that​ is, the subjects in one group cannot be related in any way to subjects in a second group. 3.The populations must be normally distributed. 4.The populations must have the same​ variance, that​ is, each treatment group has population variance σ2.

Determine the point estimate of the population​ proportion, the margin of error for the following confidence​ interval, and the number of individuals in the sample with the specified​ characteristic, x, for the sample size provided. Lower bound=0.631​, upper bound=0.859​, n=1000 a)The point estimate of the population proportion is b) The margin of error is c) The number of individuals in the sample with the specified characteristic is

a) 0.745 p-E=0.631 add P+E=0.859 2p=1.49= p=.745 b) 0.114 p+E=.859 .745+E=0.859 E=0.859-.745 E=.114 c)745 p=x/n .745=x/1000 x=.745x1000 x=745

Some have argued that throwing darts at the stock pages to decide which companies to invest in could be a successful​ stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1​ year, 52 of the companies were considered​ winners; that​ is, they outperformed other companies in the same investment class. To assess whether the​ dart-picking strategy resulted in a majority of​ winners, the researcher tested H0​:p=0.5 vs. H1​:p>0.5 & obtained a​ P-value of 0.3446. a)Explain what this​ P-value means and b)write a conclusion for the researcher.​ (Assume α is 0.1 or​ less.)

a) About 34 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. b) Because the​ P-value is​ large, do not reject the null hypothesis. There is not sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners.

Conduct a test at the α=0.10 level of significance by determining ​ (a) the null and alternative​ hypotheses, ​(b) the test​ statistic, and​ z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2 (c) the​ P-value.c) Determine the​ P-value. d)What is the result of this hypothesis​ test? Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1>p2. The sample data are x1=123​, n1=246​, x2=142​, and n2=319.

a) H0:p1=p2 H1:p1>p2 b)1.3 calculator: stats, test, (6) 2-PropzTest, p1>p2 c)0.098 d) 0.098<0.10 Reject the null hypothesis because there is sufficient evidence to conclude that p1>p2.

Test whether μ1<μ2 at the α=0.02 level of significance for the sample data shown in the accompanying table. Assume that the populations are normally distributed. n1:31 x1:103.4 s1:12.3 n2:25 x2:114.5 s2:13.2 a) Determine the null and alternative hypothesis for this test. b)Detemine the​ P-value for this hypothesis test. t=(x1-x2)/sr((s^2/n1)+(s^2/n2)) c)State the appropriate conclusion

a) H0:μ1=μ2 h1:μ1<μ2 b)p=0.001 stas: test:(4) 2-SampTtest stats pool=no c)Reject H0. There is sufficient evidence at the α=0.02 level of significance to conclude that 0.001<0.02 yes reject H0

One​ year, the mean age of an inmate on death row was 38.6 years. A sociologist wondered whether the mean age of a​ death-row inmate has changed since then. She randomly selects 32 ​death-row inmates and finds that their mean age is 37.6​, with a standard deviation of 8.9. Construct a​ 95% confidence interval about the mean age. What does the interval​ imply? a)Choose the correct hypotheses. b)Construct a​ 95% confidence interval about the mean age. E=tα/2 x (s/srn) Tα/2=1-.95/2=0.025 c)What does the interval​ imply?

a) H0=μ=38.6 H1:μ≠38.6 b)With​ 95% confidence, the mean age of a death row inmate is between 34.39 years and 40.81 years t=invT(1-.025, 31)=2.027 E=2.04x(8.9/sr32)=3.21 UB:37.6+3.21=40.81 LB:37.6-3.21=34.39 c)Since the mean age from the earlier year is contained in the​ interval, there is not sufficient evidence to conclude that the mean age had changed.

The manufacturer of a certain engine treatment claims that if you add their product to your​ engine, it will be protected from excessive wear. An infomercial claims that a woman drove 4 hours without​ oil, thanks to the engine treatment. A magazine tested engines in which they added the treatment to the motor​ oil, ran the​ engines, drained the​ oil, and then determined the time until the engines seized. a) Determine the null and alternative hypotheses that the magazine will test. b) Both engines took exactly 13 minutes to seize. What conclusion might the magazine make based on this​ evidence?

a) H0=μ=4 hours H1=μ<4 b) The​ infomercial's claim is not true.

A simple random sample of size n=15 is drawn from a population that is normally distributed. The sample mean is found to be x=22.7 and the sample standard deviation is found to be s=6.3. Determine if the population mean is different from 25 at the α=0.01 level of significance. a) Determine the null and alternative hypotheses. b) Calculate the​ P-value. t=(x-μ)/ (s/srn) 2tcdf(-99,-#,df) c) State the conclusion for the test. ​(d) State the conclusion in context of the problem.

a) H0​:μ=25 H1​:μ≠25 b)0.18 t=(22.7-25)/(6.3/sr15)=-1.41 2tcdf(-99,-1.41,14) c) 0.18>.01 Do not Reject H0 because the​P-value is greater than the α=0.01 level of significance. d. There is not sufficient evidence at the α=0.01 level of significance to conclude that the population mean is different from 25.

Suppose a researcher is testing the hypothesis H0​:p=0.4 vs H1​:p<0.4 and she finds the​ P-value to be 0.33. a)Explain what this means. Would she reject the null​ hypothesis? Why?

a) If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in about 33 of 100 samples if the null hypothesis is true. b)Since this event is not​ unusual, she will not reject the null hypothesis.

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1014 people age 15 or​ older, the mean amount of time spent eating or drinking per day is 1.34 hours with a standard deviation of 0.67 hour a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. ​(b) There are more than 200 million people nationally age 15 or older. Explain why​ this, along with the fact that the data were obtained using a random​ sample, satisfies the requirements for constructing a confidence interval c) Determine and interpret a 90​% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.E=Tα/2,df (s/srn) d) Could the interval be used to estimate the mean amount of time a​ 9-year-old spends eating and drinking each​ day? Explain.

a) Since the distribution of time spent eating and drinking each day is not normally distributed​ (skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal. b)The sample size is less than​ 5% of the population. c)1.305 to 1.375 x=1.34 s=.67 n=1014 Tα/2=(1-.90)/2=.05 invT(1-.05, 1013)=1.646 LB:1.34-1.646(.67/sr1014) UB:1.34+1.646(.67/sr1014) The nutritionist is 90​% confident that the mean amount of time spent eating or drinking per day is between 1.305 to 1.375 hours. d)No; the interval is about people age 15 or older. The mean amount of time spent eating or drinking per day for​ 9-year-olds may differ.

Sleep apnea is a disorder in which you have one or more pauses in breathing or shallow breaths while you sleep. In a​ cross-sectional study of 430 adults who suffer from sleep​ apnea, it was found that 301 had gum disease.​ Note: In the general​ population, about​ 17.5% of individuals have gum disease a) What does it mean for this study to be​ cross-sectional? b) What is the variable of interest in this​ study? Is it qualitative or​ quantitative? c) Estimate the proportion of individuals who suffer from sleep apnea who have gum disease with 95​% confidence. Interpret your result

a) The data were obtained at a specific point in time and the study was an observational study. b)Whether a person with sleep apnea has gum disease or not. Qualitative—the variable classifies the individuals in the study. c) There is 95% confidence that the population proportion of individuals who suffer from sleep apnea who have gum disease is between 0.657 & .743 p=301/430=.7 zα/2=(1-.95)/2=.025 invnorm(1-.025)=1.96 E=(1.96)sr((.7x.3)/430)=.043 .7-.043=0.657 .7+.043=0.743

In a survey of 2035 adults in a certain country conducted during a period of economic​ uncertainty, 66​% thought that wages paid to workers in industry were too low. The margin of error was 4 percentage points with 95​% confidence (a) We are 95​% confident 66​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation​ reasonable? ​(b) We are 91​% to 99​% confident 66​% of adults in the country during the period of economic uncertainty felt wages paid to workers in industry were too low. Is the interpretation​ reasonable? c) We are 95​% confident that the interval from 0.62 to 0.70 contains the true proportion of adults in the country during the period of economic uncertainty who believed wages paid to workers in industry were too low. Is the interpretation​ reasonable? d) In 95​% of samples of adults in the country during the period of economic​ uncertainty, the proportion who believed wages paid to workers in industry were too low is between 0.62 and 0.70. Is the interpretation​ reasonable?

a) The interpretation is flawed. The interpretation provides no interval about the population proportion. B)The interpretation is flawed. The interpretation indicates that the level of confidence is varying c)The interpretation is reasonable. d)The interpretation is flawed. The interpretation suggests that this interval sets the standard for all the other​ intervals, which is not true.

In​ 2003, an organization surveyed 1,500 adult Americans and asked about a certain​ war, "Do you believe the United States made the right or wrong decision to use military​ force?" Of the 1,500 adult Americans​ surveyed, 1,080 stated the United States made the right decision. In​ 2008, the organization asked the same question of 1,500 adult Americans and found that 580 believed the United States made the right decision. Construct and interpret a​ 90% confidence interval for the difference between the two population​ proportions, p2003−p2008. a)The lower bound of a​ 90% confidence interval is and The upper bound of a​ 90% confidence interval is b)Interpret the​ 90% confidence interval for the difference between the two population​ proportions, p2003−p2008.

a) The lower bound of a 90% confidence interval is 0.305 The upper bound of a 90% confidence interval is 0.361 x1=1080 n1=1500 x2=580 n2=1500 stats, test, (B) 2-Prop2int Clevel=.90 b)There is​ 90% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country is between the lower and upper bounds of the interval.

A researcher with the Department of Education followed a cohort of students who graduated from high school in a certain​ year, monitoring the progress the students made toward completing a​ bachelor's degree. One aspect of his research was to determine whether students who first attended community college took longer to attain a​ bachelor's degree than those who immediately attended and remained at a​ 4-year institution. The data in the table attached below summarize the results of his study n1=254 x1=5.49 s1=1.133 n2=1106 x2=4.57 s2=1.004 a)What is the response variable in this​ study? What is the explanatory​ variable? b) Explain why this study can be analyzed using inference of two sample means. Determine what qualifications are met to perform the hypothesis test about the difference between two means. c) Does the evidence suggest that community college transfer students take longer to attain a​ bachelor's degree? Use an α=0.10 level of significance. Perform a hypothesis test. Determine the null and alternative hypotheses. d) Determine the test statistic and p value e)Should hypothesis be rejected? d) Construct a 90​% confidence interval for μcommunity college−μno transfer to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. 2nd-vars-invt(1-α, df)= (x1-x2)+or-E e) Do the results of parts​ c) and​ d) imply that community college causes you to take extra time to earn a​ bachelor's degree?

a) The response variable is the time to graduate. The explanatory variable is the use of community college or not b)The sample sizes are not more than​ 5% of the population. The samples are independent. The samples can be reasonably assumed to be random.The sample sizes are large​ (both greater than or equal to​ 30). c)H0​: μcommunity college=μno transfer​, H1​:μcommunity college>μno transfer d)t=11.91 p=0 stas: test:(4) 2-SampTtest stats pool=no use table given e)0<0.10 Reject the null hypothesis. The evidence does suggest that community college transfer students take longer to attain a​ bachelor's degree at the α=0.10 level of significance. d)0.793 to 1.047 α=(1-.90)/2=0.05 df=smallest n-1 2nd-vars-invt(1-.05, 253)=1.651 (sr[((s^2/n1)+(s^2/n2)])=0.077 1.651x.077=.127 (5.49-4.57)+.127 (5.49-4.57)-.127 e)No observational data cannot prove causation only association.

In​ randomized, double-blind clinical trials of a new​ vaccine, were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second​ dose, 130 of 726 of subjects in the experimental group​ (group 1) experienced as a side effect. After the second​ dose, 87 of 607 of the subjects in the control group​ (group 2) experienced as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced as a side effect than subjects in group 2 at the α=0.10 level of​ significance? a)Verify the model requirements. b)Determine the null and alternative hypotheses. c)Find the test statistic for this hypothesis test. d) Pvalue e) interpret p value f)State the conclusion for this hypothesis test.

a) The samples are independent.The sample size is less than​ 5% of the population size for each sample. ed. n1p11−p1≥10 and n2p21−p2≥10 b)H0:p1=p2 vs H1:p1>p2 c)1.76 calculator: stats, test, (6) 2-PropzTest,x1=130,n1=726​,x2=87​,n2=607. p1>p2 d)0.039 under z is p in calculator or normalcdf(1.76, 9999) e)If the population are equal one would expect a sample difference proportion greater than the one observed in about 39 out of 1000 repetitions of this experiment 0.039<0.10 f)Reject H0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the α=0.10 level of significance.

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0: p=0.85 versus H1: p≠0.85 n=500, x=410, α=0.1 a)Is np01−p0≥10​? b)Now find p^ c)Find the test statistic z0= z=(p-p0)/sr ((pq)/n) d) Find the P-value: e) conclusion

a) Yes (500)(.85)(.15)=63.75 b)0.82 410/500 c)-1.88 (.82-.85)/sr((.85x.15)/500)=-1.88 d) 0.060 normalcdf(-1.88,1.88)=.940 1-.940=0.060 e)0.060<0.1 Reject the null hypothesis b/c the p value is less than α

The null and alternative hypotheses are given. a)Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. b)What parameter is being​ tested? H0​:σ=3 H1​:σ>3

a) right-tailed test b)Population standard deviation b/c the symbol is σ

A researcher wishes to compare IQ scores of patients with and without cysts. She obtains a random sample of 866 patients of each category who take an IQ test and determines each patient's IQ. a)Determine whether the following sampling is dependent or independent. b)Indicate whether the response variable is qualitative or quantitative.

a) the sampling is independent b/c an individual selected for one sample does not dictate which individual is to be in the second sample b)The variable is quantitative because it is a numerical measure.

The null and alternative hypotheses are given. a)Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. b)What parameter is being​ tested? H0​:σ=130 H1​:σ≠130

a) two-tailed test b)σ= population standard deviation

To test H0​: μ=20 versus H1​: μ<20​, a simple random sample of size n=19 is obtained from a population that is known to be normally distributed. ​(a) If x=18.2 and s=4.2​, compute the test statistic. t=(x-μ)/ (s/srn) b) graph c) Approximate the​ P-value. Choose the correct range for the​ P-value below. tcf(-99,#,df) d) If the researcher decides to test this hypothesis at the α=0.05 level of​ significance, will the researcher reject the null​ hypothesis?

a)-1.87 t=(18.2-20)/(4.2/19) b)μ<20​ so its left tailed c)0.025<p<0.05 tcd(-999,-1.87,18)=0.0389 d) The researcher will reject the null hypothesis since the P-value is less than α. is p<α? is 0.04<0.05? Yes so reject H0.

A highway safety institution conducts experiments in which cars are crashed into a fixed barrier at 40 mph. In the​ institute's 40-mph offset​ test, 40% of the total width of each vehicle strikes a barrier on the​ driver's side. The​ barrier's deformable face is made of aluminum​ honeycomb, which makes the forces in the test similar to those involved in a frontal offset crash between two vehicles of the same​ weight, each going just less than 40 mph. You are in the market to buy a family car and you want to know if the mean head injury resulting from this offset crash is the same for large family​ cars, passenger​ vans, and midsize utility vehicles​ (SUVs). The data in the accompanying table were collected from the​ institute's study. (a) State the null and alternative hypotheses. (b) Normal probability plots indicate that the sample data come from normal populations. Are the requirements to use the​ one-way ANOVA procedure​ satisfied c) Test the hypothesis that the mean head injury for each vehicle type is the same at the α=0.01 level of significance. Use technology to find the​ F-test statistic for this data set. d)Determine the​ P-value and state the appropriate conclusion below e) Draw boxplots of the three vehicle types to support the results obtained in part​s c and d

a).H0: μCars=μVans=μSUVs : at least one mean is different b)​Yes, all the requirements for use of a​ one-way ANOVA procedure are satisfied. c)0.383 add data in L1 L2 L3 stats-Test-(H) ANOVA (L1, L2, L3) L1=2nd 1 L2=2nd 2 L3=2nd 3 d)0.6869 >0.01 since the p-value is 0.6869 there is insufficient evidence to reject the null hypothesis. Thus, we cannot conclude that the means are different at the α=0.01 level of significance. e) check data to see what best fits

The research group asked the following question of individuals who earned in excess of​ $100,000 per year and those who earned less than​ $100,000 per​ year: "Do you believe that it is morally wrong for unwed women to have​ children?" Of the 1,205 individuals who earned in excess of​ $100,000 per​ year, 710 said​ yes; of the 1,310 individuals who earned less than​ $100,000 per​ year, 695 said yes. Construct a​ 95% confidence interval to determine if there is a difference in the proportion of individuals who believe it is morally wrong for unwed women to have children. the test​ statistic, z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2 a)The lower bound is b) upper bound

a)0.02 stats, test, (B) 2-Prop2int Clevel=.95 x1=710​, n1=1205​, x2=695​, and n2=1310, b)0.097 c) Because the confidence interval does not include 0, there is sufficient evidence at the α=0.05 level of significance to conclude that there is a difference in the proportions. It seems that the proportion of individuals who earn over​ $100,000 that feel it is morally wrong for unwed women to have children is greater than the proportion of individuals who earn less than​ $100,000 that feel it is morally wrong for unwed women to have children.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.67 ​hours, with a standard deviation of 2.34 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.47 ​hours, with a standard deviation of 1.83 hours. Construct and interpret a 95​% confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. a) The 95​% confidence interval for μ1−μ2 is the range from B) What is the interpretation of this confidence​ interval

a)0.25 to 2.15 α=(1-.95)/2=0.025 df=smallest n-1 2nd-vars-invt(1-.025, 39)=2.023 (sr[(s^2/n1)+(s^2/n2)])=.470 .470x2.023=0.95 (5.67-4.47)+0.95 (5.67-4.47)+0.95 b) There is 95​% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

For students who first enrolled in​ two-year public institutions in a recent​ semester, the proportion who earned a​ bachelor's degree within six years was 0.388. The president of a certain junior college believes that the proportion of students who enroll in her institution have a lower completion rate. ​(a) State the 1)null and 2)alternative hypotheses in words. ​(b) State the null and alternative hypotheses symbolically. ​(c) Explain what it would mean to make a Type I error. ​(d) Explain what it would mean to make a Type II error.

a)1)Among students who enroll at the certain junior​ college, the completion rate is 0.388. A2)Among students who enroll at the certain junior​ college, the completion rate is less than 0.388. bH0:p=.388 vs H1:p<.388 c)The president rejects the hypothesis that the proportion of studnets who earn a bachelor's degree within six years is equal to .388 ​when, in​ fact, the proportion is equal to .388. d)The president fails to rejects the hypothesis that the proportion of studnets who earn a bachelor's degree within six years is equal to .388 ​when, in​ fact, the proportion is less than .388.

9.2 Determine the​ t-value in each of the cases. ​(a) Find the​ t-value such that the area in the right tail is 0.05 with 19 degrees of freedom. b) Find the​ t-value such that the area in the right tail is 0.15 with 9 degrees of freedom. ​(c) Find the​ t-value such that the area left of the​ t-value is 0.02 with 11 degrees of freedom.​ [Hint: Use​ symmetry.] ​(d) Find the critical​ t-value that corresponds to 50​% confidence. Assume 20 degrees of freedom.

a)1.729 t19,0.05=invt(1-.05,19) b)1.100 t9,0.15=invt(1-.15,9) c)-2.328 t11,0.02=invt(0.02,11) d)0.687 1-.50=.50/2=.25 invt(1-0.25,20)

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. n=((zα/2 x s)/E)^2 a)How many subjects are needed to estimate the mean HDL cholesterol within 2 points with 99% confidence assuming s=11.7 based on earlier​ studies? b)Suppose the doctor would be content with 95% confidence. c)How does the decrease in confidence affect the sample size​ required?

a)228 E=2 s=11.7 zα/2=(1-.99)/2=.005 invnorm(1-.005)=2.576 n=((2.576x11.7)/2)^2 b)132 E=4 s=11.7 zα/2=(1-.95)/2=.025 invnorm(1-.025)=1.96 n=((1.96x11.7)/2)^2 c)Decreasing the confidence level decreases the sample size needed.

People were polled on how many books they read the previous year. Initial survey results indicate that s=10.6 books a) How many subjects are needed to estimate the mean number of books read the previous year within four books with 95​% ​confidence? ​(b) How many subjects are needed to estimate the mean number of books read the previous year within two books with 95​% ​confidence? ​(c) What effect does doubling the required accuracy have on the sample​ size? d) How many subjects are needed to estimate the mean number of books read the previous year within four books with 99​% ​confidence? Compare this result to part ​(a). How does increasing the level of confidence in the estimate affect sample​ size? Why is this​ reasonable?

a)27 E=4 s=10.6 zα/2=(1-.95)/2=.025 invnorm(1-.025)=1.96 n=((1.96x10.6)/4)^2 b)108 E=2 s=10.6 zα/2=(1-.95)/2=.025 invnorm(1-.025)=1.96 n=((1.96x10.6)/2)^2 c)Doubling the required accuracy nearly quadruples the sample size. d)47 This 99​% confidence level requires 47 subjects E=4 s=10.6 zα/2=(1-.99)/2=.005 invnorm(1-.005)=2.575 n=((2.575x10.6)/4)^2 Increasing the level of confidence increases the sample size required. For a fixed margin of​ error, greater confidence can be achieved with a larger sample size.

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a​ professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 23.1 with a standard deviation of 4.1​, while the 200 students in group 2 had a mean score of 15 with a standard deviation of 2.9. a)Determine the 90​% confidence interval for the difference in​ scores, μ1−μ2. b)Interpret the interval. c)What does this say about​ priming?

a)7.513 to 8.687 α=(1-.90)/2=0.05 df=smallest n-1 2nd-vars-invt(1-.05, 199)=1.653 (sr((s^2/n1)+(s^2/n2))=.355 .355x1.653=0.597 (23.1-15)+0.597 (23.1-15)-0.597 b)The researchers are 90​% confident that the difference of the means is in the interval. c)Since the 90​% confidence interval does not contain​ zero, the results suggest that priming does have an effect on scores.

A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. n=((zα/2 x s)/E)^2 a)How many subjects are needed to estimate the mean HDL cholesterol within 4 points with 99% confidence assuming s=13.7 based on earlier​ studies? b)Suppose the doctor would be content with 90% confidence. c)How does the decrease in confidence affect the sample size​ required?

a)78 E=4 s=13.7 zα/2=(1-.99)/2=.005 invnorm(1-.005)=2.576 n=((2.576x13.7)/4)^2 b)32 E=4 s=13.7 zα/2=(1-.90)/2=.05 invnorm(1-.05)=1.645 n=((1.645x13.7)/4)^2 c)Decreasing the confidence level decreases the sample size needed.

Assume that both populations are normally distributed. n1=19 x1=49.1 s1=6.2 n2=11 x2=41.7 s2=8.9 ​(a) Test whether μ1>μ2 at the α=0.01 level of significance for the given sample data. b) test statistics tvalue t=(x1-x2)/sr((s^2/n1)+(s^2/n2) c) Determine the critical​ value(s) (right tail so only 1) d)Should the hypothesis be​ rejected? ​(e) Construct a 959% confidence interval about μ1−μ2. 2nd-vars-invt(1-α, df)= (x1-x2)+or-E

a)H0: μ1=μ2 H1: μ1>μ2 b)2.44 stas: test:(4) 2-SampTtest stats pool=no use table given c)The critical value is 1.812. d) Reject the null hypothesis because the test statistics is in the critical region. 2.44>1.812 e)0.63 to 14.17 α=(1-.95)/2=0.025 df=smallest n-1 2nd-vars-invt(1-.025, 10)=2.228 (sr((s^2/n1)+(s^2/n2))=3.037 3.037 x 2.28=E (49.1-41.7)-E (49.1-41.7)+E

Some have argued that throwing darts at the stock pages to decide which companies to invest in could be a successful​ stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 250 companies to invest in. After 1​ year, 135 of the companies were considered​ winners; that​ is, they outperformed other companies in the same investment class. To assess whether the​ dart-picking strategy resulted in a majority of​ winners, the researcher tested H0​:p=0.5 vs. H1​:p>0.5 & obtained a​ P-value of 0.1030. a)Explain what this​ P-value means and b)write a conclusion for the researcher.​ (Assume α is 0.1 or​ less.)

a)About 10 in 100 samples will give a sample proportion as high or higher than the one obtained if the population proportion really is 0.5. b) Because the​ P-value is​ large, do not reject the null hypothesis. There is not sufficient evidence to conclude that the​ dart-picking strategy resulted in a majority of winners.

Twenty years​ ago, 55​% of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 255 of 700 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did twenty years​ ago? Use the α=0.01 level of significance. a)np0(1−p0)=(700)(.55)(1-.55) b)What are the null and alternative​ hypotheses? c)Find the test​ statistic, z0. z=(p-p0)/sr ((pq)/n) d) Find the P-value: e) conclusion

a)Because np0(1−p0)=​173.3>10,the sample size is less than ​5% of the population​ size, and the sample can be reasonably assumed to be random, the requirements for testing the hypothesis are satisfied b)H0:p=0.55 H1: p≠0.55 c)-9.89 255/700=.364 (.364-.55)/sr((.55x.45)/700)=-9.89 d) 0.092 normalcdf(-9.89,9.89)=1 1-1=0 e)0<0.01 Since ​P-value < α​, reject the null hypothesis and conclude that there is not sufficient evidence that parents feel differently today.

In a clinical​ trial, 21 out of 863 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.2​% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.2​%of this​ drug's users experience flulike symptoms as a side effect at the α=0.01 level of​ significance? a)np0(1−p0)=(863)(.022)(1-.022) b)What are the null and alternative​ hypotheses? c)Find the test​ statistic, z0. z=(p-p0)/sr ((pq)/n) d) Find the P-value: e) conclusion

a)Because np0(1−p0)=​18.6>10,the sample size is less than ​5% of the population​ size, and the sample can be reasonablys assumed to be random the requirements for testing the hypothesis are satisfied b)H0:p=0.022 H1: p>0.022 c)0.40 21/863=.024 (.024-.022)/sr((.022x.978)/863)=0.40 d) .345 normalcdf(0.4,99)=.345 e)0.345>0.01 Since ​P-value>α​, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.2​% of the users experience flulike symptoms

In a previous​ poll, 31​% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose​ that, in a more recent​ poll, 1166 adults with children under the age of 18 were selected at​ random, and 340 of those 1166 adults reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ decreased? Use the α=0.01 significance level. a)np0(1−p0)=(1166)(.31)(1-.31) b)What are the null and alternative​ hypotheses? c)Find the test​ statistic, z0. z=(p-p0)/sr ((pq)/n) d) Find the P-value: e) conclusion

a)Because np0(1−p0)=​249.4>10,the sample size is less than ​5% of the population​ size, and the adults in the sammple were selected at random, all of the requirements for testing the hypothesis are satisfied b)H0:p=0.31 H1: p<.31 c)-1.33 340/1166=.292 (.292-.31)/sr((.31x.69)/1166)=-1.33 d) 0.092 normalcdf(-999,-1.33) e)0.092>0.01 Since ​P-value greater than α​, do not reject the null hypothesis. There is not sufficient evidence at the α​=0.01 level of significance to conclude that the proportion of families with children under the age of 18 who eat together seven nights a week is less than 0.31.

In a​ survey, 44​% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too​ high, so she randomly selected 210 pet owners and discovered that 87 of them spoke to their pet on the telephone. Does the veterinarian have a right to be​ skeptical? Use the α=0.01 level of significance. a)np0(1−p0)=(210)(.44)(1-.44) b)What are the null and alternative​ hypotheses? c)Find the test​ statistic, z0. z=(p-p0)/sr ((pq)/n) d) Find the P-value: e) conclusion

a)Because np0(1−p0)=​51.7>10,the sample size is less than ​5% of the population​ size, and the sample is given to be random, the requirements for testing the hypothesis are satisfied b)H0:p=0.44 H1: p<0.44 c)-0.76 87/210=.414 (.414-.44)/sr((.44x.56)/210)=-0.76 d) 0.224 normalcdf(-999,-.76)=.224 e)0.224>0.01 The veterinarian does not have a right to be skeptical. There is not sufficient evidence to conclude that the true proportion of pet owners who talk to their pets on the telephone is less than 44​%.

Researchers wanted to evaluate whether​ ginkgo, an​ over-the-counter herb marketed as enhancing​ memory, improves memory in elderly adults as measured by objective tests. To do​ this, they recruited 97 men and 125 women older than 55 years and in good health. Participants were randomly assigned to receive​ ginkgo, 45 milligrams​ (mg) 3 times per​ day, or a matching placebo. The measure of memory improvement was determined by a standardized test of learning and memory. After 6 weeks of​ treatment, the data indicated that ginkgo did not increase performance on standard tests of​ learning, memory,​ attention, and concentration. These data suggest​ that, when taken following the​ manufacturer's instructions, ginkgo provides no measurable increase in memory or related cognitive function to adults with healthy cognitive function. a) What type of experimental design is​ this? b)What is the population being​ studied? ​(c) What is the response variable in this​ study? d)What is the factor that is set to predetermined​ levels? e)What are the​ treatments? f)Identify the experimental units g)What is the control group in this​ study?

a)Completely randomized design b)Adults older than 55 years and in good health c)The score on the standardized test of learning and memory d) the drug: 45 mg 3 times a day or a matching placebo e)45 mg of the drug 3 times a day or a matching placebo f)The 97 men and 125 women older than 55 who are in good health that participated in the study g)The placebo group serves as the control group because this group corresponds to the reference point that will be compared to the other group.

Assume that both populations are normally distributed. ​(a) Test whether μ1≠μ2 at the α=0.01 level of significance for the given sample data. b) pvalue c) Conclusion ​(d) Construct a 959% confidence interval about μ1−μ2. 2nd-vars-invt(1-α, df)= (x1-x2)+or-E

a)H0: μ1=μ2 H1: μ1≠μ2 b)0.001 stas: test:(4) 2-SampTtest stats pool=no use table given c)0.001<0.01 Reject H0​, there is sufficient evidence to conclude that the two populations have different means. d)-7.9 to -0.70 α=(1-.99)/2=0.005 df=smallest n-1 2nd-vars-invt(1-.005, 15)=2.947 (sr((s^2/n1)+(s^2/n2))=# # x 2.947=E (12.6-16.9)-E (182.6-16.9)+E

Conduct a test at the α=0.01 level of significance. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1>p2. The sample data are x1=122​, n1=254​, x2=134​, and n2=317. ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2 c) the​ P-value. d)What is the result of this hypothesis​ test?

a)H0:p1=p2 vs H1:p1>p2 b)1.38 calculator: stats, test, (6) 2-PropzTest, x1=122​, n1=254​, x2=134​, and n2=317, p1>p2 c)0.085 or under z is p in calculator normalcdf(1.38, 99) d)Do not reject the null hypothesis b/c there is not sufficient evidence to conclude that p1>p2 0.085>0.01

Conduct a test at the α=0.05 level of significance. Assume the samples were obtained independently from a large population using simple random sampling.Test whether p1≠p2.Sample data are x1=30​,n1=254​,x2=38​,n2=301. ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, z0= (p1-p2)/( (srpq)(1/n1 +1/n2)) p=(x1+x2)/(n1+n2) p1=x1/n1 p2=x2/n2 c) the​ P-value. d)What is the result of this hypothesis​ test?

a)H0:p1=p2 vs H1:p1≠p2 b)-0.29 calculator: stats, test, (6) 2-PropzTest, x1=30​,n1=254​,x2=38​,n2=301. p1≠p2 c)0.771 or under z is p in calculator 2normalcdf(-999,-.29) d)Do not reject the null hypothesis b/c there is not sufficient evidence to conclude thatp1≠p2 0.771>0.05

In 1945​, an organization asked 1457 randomly sampled American​ citizens, "Do you think we can develop a way to protect ourselves from atomic bombs in case others tried to use them against​ us?" with 763 responding yes. Did a majority of the citizens feel the country could develop a way to protect itself from atomic bombs in 1945​?Use the α=0.01 level of significance. A) What are the null and alternative​ hypotheses? b)Determine the test​ statistic, z0. c)Use technology to determine the​ P-value for the test statistic. d)What is the correct conclusion at the α=0.01 level of​ significance?

a)H0:p=0.5 vs H1:p>0.5 b)1.81 763/1457=.5237 (.5237-.5)/sr((.5x.5)/1457) c)0.035 normalcdf(1.81,999) d) Since the​ P-value is greater than the level of​ significance, do not reject the null hypothesis. There is not sufficient evidence to conclude that the majority of the citizens feel the country could develop a way to protect itself from atomic bombs.

The average daily volume of a computer stock in 2011 was μ=35.1 million​ shares, according to a reliable source. A stock analyst believes that the stock volume in 2018 is different from the 2011 level. Based on a random sample of 30 trading days in​ 2018, he finds the sample mean to be 26.8 million​ shares, with a standard deviation of s=11.8 million shares. Test the hypotheses by constructing a 95​% confidence interval. ​(a) State the hypotheses for the test b) Construct a 95​% confidence interval about the sample mean of stocks traded in 2018. E=tα/2 x (s/srn) Tα/2=1-.95/2=0.025 c) Will the researcher reject the null​ hypothesis?

a)H0:μ=35.1million​ sharees H1:μ≠35.1million​ sharees b)With 95​%​confidence, the mean stock volume in 2018 is between 22.394 million shares and 31.206 million shares. t=invT(1-.025, 29)=2.0452 E=2.0452x(11.8/sr30)=4.40613 UB:26.8+4.4061=31.206 LB:26.8-4.4061=22.394 c)Reject the null hypothesis because μ=35.1 million shares does not fall in the confidence interval.

A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the α=0.01 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. a)State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms. b) p value the numbers are data so add to L1 and L2 c) conclusion

a)H0​:μ1=μ2 H1​:μ1>μ2 b)p=0.233 add data to L1 and L2 stas: test:(4) 2-SampTtest data pool=no fre=1 c)0.233>0.01 Do not reject H0. There is not significant evidence at the α=0.01 level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

A college entrance exam company determined that a score of 21 on the mathematics portion of the exam suggests that a student is ready for​ college-level mathematics. To achieve this​ goal, the company recommends that students take a core curriculum of math courses in high school. Suppose a random sample of 150 students who completed this core set of courses results in a mean math score of 21.3 on the college entrance exam with a standard deviation of 3.7. Do these results suggest that students who complete the core curriculum are ready for​ college-level mathematics? That​ is, are they scoring above 21 on the mathematics portion of the​ exam? α=0.10 level of significance to test a) State the appropriate null and alternative hypotheses. ​b) Verify that the requirements to perform the test using the​ t-distribution are satisfied. c) Test statistics d) Pvalue e)Conclusion

a)H0​:μ=21 vs H1:μ>21 b)The students were randomly sampled. The sample size is larger than 30. The​ students' test scores were independent of one another. c)0.99 (21.3-21)/(3.7sr150) d)0.162 tcdf(0.99,999,249) e)Do not reject the null hypothesis and claim that there is not sufficient evidence to conclude that the population mean is greater than 21.

A television sports commentator wants to estimate the proportion of citizens who​ "follow professional​ football." (a) What sample size should be obtained if he wants to be within 4 percentage points with 94​% confidence if he uses an estimate of 54​% obtained from a​ poll? (b) What sample size should be obtained if he wants to be within 4 percentage points with 94​% confidence if he does not use any prior​ estimates? c) Why are the results from parts​ (a) and​ (b) so​ close?

a)sample size=549 zα/2=(1-.94)=.06 .06/2=0.03=1-.03=.97 invonorm(.97)=1.88 n=p^q^((zα/2)/E)^2 p=.54 q=.46 E=.04 zα/2=1.88 n=(.54x.46)(1.88/.04)^2=549 B) sample size=553 n=.25((zα/2)/E)^2 n=.25(1.88/.04)^2 c)The results are close because 0.54(1−0.54)=0.2484 is very close to 0.25

Suppose the mean​ wait-time for a telephone reservation agent at a large airline is 43 seconds. A manager with the airline is concerned that business may be lost due to customers having to wait too long for an agent. To address this​ concern, the manager develops new airline reservation policies that are intended to reduce the amount of time an agent needs to spend with each customer. A random sample of 250 customers results in a sample mean​ wait-time of 42.3 seconds with a standard deviation of 4.2 seconds. Using α=0.05 level of​ significance, do you believe the new policies were effective in reducing wait​ time? Do you think the results have any practical​ significance? A)Determine the null and alternative hypotheses. b)Calculate the test statistic. t0 c)Calculate the​ P-value. d)State the conclusion for the test. State the conclusion in context of the problem. e)Do you think the results have any practical​ significance?

a)H0​:μ=43 sec vs H1:μ<43 sec b)-2.64 (42.3-43)/(4.2sr250) c)0.004 tcdf(-999,-2.64,249) d)H0 because the​P-value is less than the α=0.05 level of significance. There is sufficient evidence at the α=0.05 level of significance to conclude that the new policies were effective. e)No, because while there is significant evidence that shows the new policies were effective in lowering the mean​ wait-time of​ customers, the difference between the previous mean​ wait-time and the new mean​ wait-time is not large enough to be considered important.

According to a food​ website, the mean consumption of popcorn annually by Americans is 60 quarts. The marketing division of the food website unleashes an aggressive campaign designed to get Americans to consume even more popcorn. a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the marketing campaign. b) A sample of 871 Americans provides enough evidence to conclude that marketing campaign was effective. Provide a statement that should be put out by the marketing department. c)​ Suppose, in​ fact, the mean annual consumption of popcorn after the marketing campaign is 60 quarts. Has a Type I or Type II error been made by the marketing​ department? If we tested this hypothesis at the α=0.01 level of​ significance, what is the probability of committing this​ error?

a)H0​:μ=60 H1​:μ>60 b)There is sufficient evidence to conclude that the mean consumption of popcorn has risen. c)The marketing department committed a Type I error because the marketing department rejected the null hypothesis when it was true. The probability of making a Type I error is 0.01

Several years​ ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 64.5 inches. ​(a) State the appropriate null and alternative hypotheses to assess whether women are taller today. ​(b) Suppose the​ P-value for this test is 0.04. Explain what this value represents. ​(c) Write a conclusion for this hypothesis test assuming an α=0.10 level of significance.

a)H0​:μ=63.7 in. vs. H1:μ>63.7in. b)There is a 0.04 probability of obtaining a sample mean height of 64.5 inches or taller from a population whose mean height is 63.7 inches. c)Reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today Is 0.04<0.10? Yes so reject H0

It has long been stated that the mean temperature of humans is 98.6°F. ​However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98.6°F. They measured the temperatures of 61 healthy adults 1 to 4 times daily for 3​ days, obtaining 275 measurements. The sample data resulted in a sample mean of 98.4°F and a sample standard deviation of 1°F. Use the​ P-value approach to conduct a hypothesis test to judge whether the mean temperature of humans is less than 98.6°F at the α=0.01 level of significance. a)State the hypotheses. b)Identify the​ t-statistic. t=(x-μ)/ (s/srn) c) Identify the​ P-value.μ<left:tcf(-99,#,df) d)What can be​ concluded?

a)H0​:μ=98.6°F H1​:μ<98.6°F b)-3.32 t=(98.4-98.6)/(1/sr275) c)0.001 tcdf(-999,-3.32,274)=0.0005 d)Reject H0 since the​ P-value is less than the significance level. Is p<α? 0.001<0.01? Yes reject H0

A psychologist obtains a random sample of 20 mothers in the first trimester of their pregnancy. The mothers are asked to play Mozart in the house at least 30 minutes each day until they give birth. After 5​ years, the child is administered an IQ test. It is known that IQs are normally distributed with a mean of 100. If the IQs of the 20 children in the study result in a sample mean of 104.5 and sample standard deviation of 14.2​, is there evidence that the children have higher​ IQs? Use the α=0.10 level of significance. a) What type of hypothesis test is appropriate to conduct for this​ research? b) Determine the null and alternative hypotheses. c)Which distribution should be used for this hypothesis​ test? d)Calculate the test statistic. t=(x-μ)/ (s/srn) t=104.5 μ=100 s=14.2 n=20 e) p value tcf(#T,99999,df) d)State the conclusion for the test

a)Hypothesis test on a population mean b)H0​:μ=100 H1​:μ>100 c)The​ Student's t-distribution because the parameter is the​ mean, σ is not​ known, and the model conditions are satisfied. d)1.42 (104.5-100)/(14.2/sr20) e)0.086 tcf(1.42,9999,19) d) 0.086<0.10 Reject H0. There is sufficient evidence at the α=0.10 level of significance to conclude that mothers who listen to Mozart have children with higher IQs.

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean, x​, is found to be 110​, and the sample standard​ deviation, s, is found to be 10. x=110 Tα=1-% df=n-1 E=Tα/2,df (s/srn) ​(a) Construct a 95% confidence interval about μ if the sample​ size, n, is 22. ​(b) Construct a 95% confidence interval about μ if the sample​ size, n, is 15. How does decreasing the sample size affect the margin of​ error, E? ​(c) Construct a 80​% confidence interval about μ if the sample​ size, n, is 22. Compare the results to those obtained in part​ (a). How does decreasing the level of confidence affect the size of the margin of​ error, E? ​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

a)Lb=105.6 ub=114.4 Tα/2=(1-.95)/2=.025 invT(1-.025, 21)=2.08 LB:110-2.08(10/sr22) UB:110+2.08(10/sr22) b)104.5-115.5 Tα/2=(1-.95)/2=.025 invT(1-.025, 14)=2.131 LB:110-2.131(10/sr15) UB:110+2.131(10/sr15) As the sample size decreases​,the margin of error increases. c)107.2 to 112.8 Tα/2=(1-.8)/2=.10 invT(1-.10, 21)=1.323 LB:110-1.323(10/sr22) UB:110+1.323(10/sr22) As the level of confidence decreases​, the size of the interval decreases. d) No, the population needs to be normally distributed.

A simple random sample of size n is drawn. The sample​ mean, x​, is found to be 18.7​, and the sample standard​ deviation, s, is found to be 4.4. a) Construct a 95​% confidence interval about μ if the sample​ size, n, is 35. b)Construct a 95​% confidence interval about μ if the sample​ size, n, is 51.How does increasing the sample size affect the margin of​ error, E? c)Construct a 99​% confidence interval about μ if the sample​ size, n, is 35 Compare the results to those obtained in part​ (a). How does increasing the level of confidence affect the size of the margin of​ error, E? d)) If the sample size is 13​, what conditions must be satisfied to compute the confidence​ interval?

a)Lb=17.19 ub=20.21 Tα/2=(1-.95)/2=.025 invT(1-.025, 34)=2.032 LB:18.7-2.032(4.4/sr35) UB:18.7+2.032(4.4/sr35) b)Lb=17.46 ub=19.94 Tα/2=(1-.95)/2=.025 invT(1-.025, 50)=2.009 LB:18.7-2.009(4.4/sr51) UB:18.7+2.009(4.4/sr51) The margin of error decreases. c)Lb=16.67 ub=20.73 Tα/2=(1-.99)/2=.005 invT(1-.005, 34)=2.728 LB:18.7-2.728(4.4/sr35) UB:18.7+2.728(4.4/sr35) The margin of error increases. d)The sample data must come from a population that is normally distributed with no outliers.

The standard deviation in the pressure required to open a certain valve is known to be σ=1.3 psi. Due to changes in the manufacturing​ process, the​ quality-control manager feels that the pressure variability has been reduced. ​(a) State the null and alternative hypotheses in words. ​(b) State the null and alternative hypotheses symbolically. ​(c) Explain what it would mean to make a Type I error. ​(d) Explain what it would mean to make a Type II error.

a)Null: The standard deviation in the pressure required to open a certain valve is 1.3 psi. alternative: The standard deviation in the pressure required to open a certain valve is less than 1.3 psi. b) H0: σ=1.3 psi H1: σ<1.3 psi c) The manager rejects the hypothesis that the pressure variability is equal to 1.3 ​psi, when the true pressure variability is equal to 1.3 psi. d)The manager fails to rejects the hypothesis that the pressure variability is equal to 1.3 ​psi, when the true pressure variability is less than 1.3 psi.

A simple random sample of size n=200 drivers with a valid​ driver's license is asked if they drive an​ American-made automobile. Of the 200 drivers​ surveyed, 108 responded that they drive an​ American-made automobile. Determine if a majority of those with a valid​ driver's license drive an​ American-made automobile at the α=0.05 level of significance. a) What type of variable is​ "drive an​ American-made automobile, or​ not"? b)What type of hypothesis test is appropriate to conduct for this​ research? c)Determine the null and alternative hypotheses. Which distribution should be used for this hypothesis​ test? d)Calculate the test statistic. e)Calculate the​ P-value. f)State the conclusion for the test.

a)Qualitative with two possible outcomes b)Hypothesis test on a population proportion c)H0:p=.5 vs H1:p>.5 The normal distribution because the parameter is a​ proportion, p, and the model conditions are satisfied. d)1.13 108/200=.54 (.54-.5)/sr((.5x.5)/200) e)0.129 normalcdf(1.13,999) f)Do not reject H0. There is not sufficient evidence at the α=0.05 level of significance to conclude that more than half of all drivers with a valid​ driver's license drive an​ American-made automobile.

A simple random sample of size n=200 drivers with a valid​ driver's license is asked if they drive an​ American-made automobile. Of the 200 drivers​ surveyed, 121 responded that they drive an​ American-made automobile. Determine if a majority of those with a valid​ driver's license drive an​ American-made automobile at the α=0.05 level of significance. a) What type of variable is​ "drive an​ American-made automobile, or​ not"? b)What type of hypothesis test is appropriate to conduct for this​ research? c)Determine the null and alternative hypotheses. Which distribution should be used for this hypothesis​ test? d)Calculate the test statistic. e)Calculate the​ P-value. f)State the conclusion for the test.

a)Qualitative with two possible outcomes b)Hypothesis test on a population proportion c)H0:p=.5 vs H1:p>.5 The normal distribution because the parameter is a​ proportion, p, and the model conditions are satisfied. d)2.97 121/200=.605 (.605-.5)/sr((.5x.5)/200) e)0.001 normalcdf(2.97,999) f)Reject H0. There is sufficient evidence at the α=0.05 level of significance to conclude that more than half of all drivers with a valid​ driver's license drive an​ American-made automobile.

Three years​ ago, the mean price of an existing​ single-family home was ​$243,700. A real estate broker believes that existing home prices in her neighborhood are higher. ​(a) State the null and alternative hypotheses in words. ​(b) State the null and alternative hypotheses symbolically. ​(c) Explain what it would mean to make a Type I error. ​(d) Explain what it would mean to make a Type II error.

a)The mean price of a single family home in the​ broker's neighborhood is ​$243,700. alternative: The mean price of a single family home in the​ broker's neighborhood is greater than ​$243,700. b)H0=μ=243700 H1=μ>243700 c) The broker rejects the hypothesis that the mean price is equal to $243700 when the true mean price is equal to $243700 d)The broker fails to rejects the hypothesis that the mean price is equal to $243700 when the true mean price is greater than $243700

In 1940​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alcoholic​ beverages?" Of the 1100 adults​ surveyed, 396 indicated that they were total abstainers. In a recent​ survey, the same question was asked of 1100 adults and 330 indicated that they were total abstainers a) Determine the sample proportion for each sample.p1=x1/n1 p2=x2/n2 b)First verify the model requirements c)Determine the null and alternative hypotheses. d)Find the test statistic for this hypothesis test. e) Pvalue f) interpret p value g)State the conclusion for this hypothesis test.

a)The proportions of the adults who took the 1940 survey and the recent survey who were total abstainers are 0.36 and 0.30 respectively p1=396/1100 p2=330/1100 B)The samples are independent.The sample size is less than​ 5% of the population size for each sample. n1p11−p1≥10 and n2p21−p2≥10 c)H0​:p1=p2 H1​:p1≠p2 d)2.99 calculator: stats, test, (6) 2-PropzTest,x1=396,n1=1100​,x2=330​,n2=1100. p1≠p2 e)0.003 under z is p in calculator f)If the population are equal one would expect a sample difference proportion greater than the absolute value of the one observed in about 0 out of 100 repetitions of this experiment g)0.003<0.10 Reject H0. There is sufficient evidence at that =0.10 level of significance to suggest that proportion of adults who totally abstain from alcohol has changed

Do people walk faster in the airport when they are departing​ (getting on a​ plane) or do they walk faster when they are arriving​ (getting off a​ plane)? A reputable researcher measured the walking speed of random travelers in two International Airports. His findings are summarized in the table. a)Is this an observational study or a designed​ experiment? Why? x=mean s=standard deviation and sample size is n b) Explain why it is reasonable to use​ Welch's t-test. c) Do individuals walk at different speeds depending on whether they are departing or arriving at the α=0.05 level of​ significance? Let μ1 represent the mean speed of people departing and μ2 represent the mean speed of people arriving. d)Determine the​ P-value for this hypothesis test. e)correct conclusion?

a)This is an observational study since the researcher did not influence the data. b)The samples are​ random, large and independent. c)h0:μ1=μ2 H1​:μ1≠μ2 d)0.431 stas: test:(4) 2-SampTtest stats pool=no use table given e)0.431>0.05 Do not reject H0. There is not sufficient evidence at the α=0.05 level of significance to say that travelers walk at different speeds depending on whether they are arriving or departing.

An educator wants to determine whether a new curriculum significantly improves standardized test scores for third grade students. She randomly divides 100 third-graders into two groups. Group 1 is taught using the traditional ​curriculum, while group 2 is taught using the new curriculum. At the end of the school​ year, both groups are given the standardized test and the mean scores are compared. a)Determine whether the sampling is dependent or independent. b)Indicate whether the response variable is qualitative or quantitative

a)This sampling is independent because the individuals selected for one sample do not dictate which individuals are to be in a second sample. b)The variable is quantitative because it is a numerical measure.

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0: p=0.52 versus H1: p<0.52 n=150, x=66, α=0.05 p=66/150=.44 a) Is np0(1−p0)≥10​? b) ​P-value? z=(p-p0)/sr ((pq)/n) c) do you reject H0?

a)Yes (150)(.52)(.48)=37.44 b)0.025 (.44-.52)/sr((.52x.48)/150)=-1.961 normalcdf(-999,-1.961) c).025<.05: Reject the null hypothesis b/c the p value is less than α

In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsible for​ long-term memory​ storage, in adolescents. The researchers randomly selected 20 adolescents with alcohol use disorders to determine whether the hippocampal volumes in the alcoholic adolescents were less than the normal volume of 9.02 cm^3. An analysis of the sample data revealed that the hippocampal volume is approximately normal with no outliers and x=8.17 cm^3 and s=0.7 cm^3. Conduct the appropriate test at the α=0.01 level of significance. a) State the null and alternative hypotheses. b) Identify the​ t-statistic. t=(x-μ)/ (s/srn) c) Identify the​ P-value.μ<left:tcf(-99,#,df) d)Make a conclusion regarding the hypothesis.

a)h0:​μ=9.02 h1=μ<9.02 b)-5.43 (8.17-9.02)/(.7/sr20) c)0 tcdf(-999,-5.43,19)=0.000015 d)Reject the null hypothesis. There is sufficient evidence to claim that the mean hippocampal volume is less than 9.02 cm^3. Is 0<0.01? yes reject

A physical therapist wants to determine the difference in the proportion of men and women who participate in regular sustained physical activity. What sample size should be obtained if she wishes the estimate to be within three percentage points with 99​% ​confidence, assuming that (0.03 is from within three percentage) ​(a) she uses the estimates of 21.7​% male and 18.3​% female from a previous​ year? n=[p1(1-p1) +p2(1-p2)]x[(zα/2)/E]^2 ​(b) she does not use any prior​ estimates? n=0.5[(zα/2)/E]^2

a)n=2356 p1=.217 p2=.183 E=.03 zα/2=1-.99/2=.005=2.576 invnorm 1-.005=2.576 (.217x(1-.217)+.183(1-.183))x(2.576/0.03)^2=2356 b)3687 0.5(2.576/0.03)^2

A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 5 percentage points with 95​% confidence if ​(a) he uses a previous estimate of 26​%? ​(b) he does not use any prior​ estimates?

a)n=296 n=p^q^((zα/2)/E)^2 p=.26 q=.74 E=.05 zα/2=95%=0.025 invnorm(1-.025)=1.960 n=(.26x.74)(1.96/.05)^2=1183 b)n=385 n=.25((zα/2)/E)^2 n=.25(1.96/.05)^2

A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 5 percentage points with 95​% confidence if ​(a) he uses a previous estimate of 28​%? ​(b) he does not use any prior​ estimates?

a)n=412 n=p^q^((zα/2)/E)^2 p=.28 q=.72 E=.05 zα/2=.95%=1.96 n=(.28x.72)(1.96/.05)^2=310 b)n=385 n=.25((zα/2)/E)^2 n=.25(1.96/.05)^2

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.04 with 90​% confidence if ​(a) she uses a previous estimate of 0.42​? ​(b) she does not use any prior​ estimates?

a)n=412 n=p^q^((zα/2)/E)^2 p=.42 q=.58 E=.04 zα/2=.90%=0.5 invnorm(1-.05)=1.645 n=(.42x.58)(1.645/.04)^2=412 b)n=423 n=.25((zα/2)/E)^2 n=.25(1.645/.04)^2

According to a​ report, the standard deviation of monthly cell phone bills was ​$4.87 in 2017. A researcher suspects that the standard deviation of monthly cell phone bills is different today. ​(a) State the null and alternative hypotheses in words. ​(b) State the null and alternative hypotheses symbolically. ​(c) Explain what it would mean to make a Type I error. ​(d) Explain what it would mean to make a Type II error.

a)null:The standard deviation of monthly cell phone bills is $4.87. alter: :The standard deviation of monthly cell phone bills is different from $4.87 b) H0: σ=$ 4.87 H1: σ≠$4.87 c)The sample evidence led the researcher to believe the standard deviation of monthly cell phone bills is different from ​$4.87 ​when, in​ fact, the standard deviation of bills is $4.87. d)The sample evidence did not lead the researcher to believe the standard deviation of monthly cell phone bills is different from ​$4.87 ​when, in​ fact, the standard deviation of bills is different from $4.87.

The null and alternative hypotheses are given. a)Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. B)What parameter is being​ tested? H0​:p=0.4 H1​:p<0.4

a)p<=left-tailed test B) P= Population proportion b/c symbol used in hypotheses is p

The null and alternative hypotheses are given. a)Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. b)What parameter is being​ tested? H0​:p=0.8 H1​:p>0.8

a)p> is right-tailed test b) Population proportion b/c symbol used in hypotheses is p

To test H0: μ=103 VS. H1: μ≠103 a simple random sample of size n=35 is obtained. a) Does the population have to be normally distributed to test this​ hypothesis? Why? b)If x=99.9 and s=5.9​, compute the test statistic. t=(x-μ)/ (s/srn) c) graph d)Approximate the​ P-value. Choose the correct range for the​ P-value below. 2tcdf(-99,-#,df) Interpret the​ P-value. e) If the researcher decides to test this hypothesis at the α=0.05 level of​ significance, will the researcher reject the null​ hypothesis?

a)​​No, because n≥30. b)-3.11 t=(99.9-103)/(5.9/35) c) μ≠103 means 2 tailed d) 0.002<​P-value<0.005 p=tcdf(-999,-3.11,34)=.001886x2=0.00377 If 1000 random samples of size n=35 are​ obtained, about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if μ=103. e) Yes is p<α? is 0.004<0.05? yes reject

The data found below measure the amounts of greenhouse gas emissions from three types of vehicles. The measurements are in tons per​ year, expressed as CO2 equivalents. Use a 0.025 significance level to test the claim that the different types of vehicle have the same mean amount of greenhouse gas emissions. Based on the​ results, does the type of vehicle appear to affect the amount of greenhouse gas​ emissions? data on table a. What are the hypotheses for this​ test? b. Determine the test statistic. F value c. Pvalue d. What is the conclusion of the​ test?

a. H0=μ1=μ2=μ3 H1​: At least one of the means is different from the others b. F=1.42 add data in L1 L2 L3 stats-Test-(H) ANOVA (L1, L2, L3) L1=2nd 1 L2=2nd 2 L3=2nd 3 c. p=0.26 d. Do not reject the null hypothesis. Conclude that the type of vehicle does not appear to affect the amount of greenhouse gas emissions for these three types. 0.26>0.025

Refer to the accompanying data​ table, which shows the amounts of nicotine​ (mg per​ cigarette) in​ king-size cigarettes,​ 100-mm menthol​ cigarettes, and​ 100-mm nonmenthol cigarettes. The​ king-size cigarettes are​ nonfiltered, while the​ 100-mm menthol cigarettes and the​ 100-mm nonmenthol cigarettes are filtered. Use a 0.05 significance level to test the claim that the three categories of cigarettes yield the same mean amount of nicotine. Given that only the​ king-size cigarettes are not​ filtered, do the filters appear to make a​ difference? a. Determine the null and alternative hypotheses. b. Determine the test statistic. F value c. Pvalue d. What is the conclusion of the​ test? e. Do the filters appear to make a​ difference?

a. H0=μ1=μ2=μ3 H1​: At least one of the three population means is different from the others b. F=4.6259 add data in L1 L2 L3 L4 stats-Test-(H) ANOVA (L1, L2, L3, L4) L1=2nd 1 L2=2nd 2 L3=2nd 3 c. 0.0187 d. 0.0187< 0.05 Reject H0. There is sufficient evidence to warrant rejection of the claim that the three categories of cigarettes yield the same mean amount of nicotine. e. Given that the king-size cigarettes have the largest mean, it appears that the filters do make a difference (although this conclusion is not justified by the results from analysis of variance

Weights​ (kg) of poplar trees were obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the accompanying table. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most​ effective? a. Determine the null and alternative hypotheses. b. Determine the test statistic. F value c. Pvalue d. What is the conclusion of the​ test? e. Is there a treatment that appears to be most​ effective?

a. H0=μ1=μ2=μ3=μ4 H1​: At least one of the four population means is different from the others b. F=3.6292 add data in L1 L2 L3 L4 stats-Test-(H) ANOVA (L1, L2, L3, L4) L1=2nd 1 L2=2nd 2 L3=2nd 3 c. p=0.0360 d. 0.036<0.05 Reject H0. There is sufficient evidence to warrant rejection of the claim that the four different treatments yield the same mean poplar weight. e. The 'Fertilizer and Irrigation' method seems to be most effective. Had bigger numbers

A certain statistics instructor participates in triathlons. The accompanying table lists times​ (in minutes and​ seconds) he recorded while riding a bicycle for five laps through each mile of a​ 3-mile loop. Use a 0.05 significance level to test the claim that it takes the same time to ride each of the miles. Does one of the miles appear to have a​ hill? mile1: 3.15 3.25 3.23 3.23 3.22 mile2: 3.18 3.23 3.21 3.18 3.19 mile3: 3.35 3.32 3.30 3.32 3.28 a. Determine the null and alternative hypotheses. b. Find the F test statistic. c. Pvalue d. What is the conclusion for this hypothesis​ test? e. Does one of the miles appear to have a​ hill?

a. H0​:μ1=μ2=μ3 H1= At least one of the 3 population means is different from the others b. F=22.2281 add data in L1 L2 L3 stats-Test-(H) ANOVA (L1, L2, L3) L1=2nd 1 L2=2nd 2 L3=2nd 3 c. 0.0001 d. Reject H0. There is sufficient evidence to warrant rejection of the claim that the three different miles have the same mean ride time. 0.0001<0.05 e. Yes these data suggest that the third mile appears to take longer and a reasonable explanation is that is has a hill.

Samples of pages were randomly selected from three different novels. The Flesch Reading Ease scores were obtained from each​ page, and the​ TI-83/84 Plus calculator results from analysis of variance are given below. Use a 0.01 significance level to test the claim that the three books have the same mean Flesch Reading Ease score. One-way ANOVA F=9.5673184393 p=6.1094562E−4 Factor df=2 SS=1369.42574 ↓ MS=684.712871 ​One-way ANOVA ↑ MS=684.712871 Error df=30 SS=2147.03694 MS=71.5678981 Sxp=8.45978121 pop-up content endsPrintDone a. What is the conclusion for this hypothesis​ test?

a. Reject H0. There is sufficient evidence to warrant rejection of the claim that the 3 books have the same mean Flesch Reading Ease score. p=6.1094562E−4<0.01

A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions are approximately normal and the variances do not differ greatly. Use a 0.05 significance level to test the claim that the mean weight of different colored candies is the same. If the candy maker wants the different color populations to have the same mean​ weight, do these results suggest that the company has a problem requiring corrective​ action? ​Source: ​Treatment: df=7 SS=0.266 MS=0.038 Test stat F=4.2776 Critical F=2.1263 pvalue=0.0005 Error: df=80 ss=0.720 ms=0.009 total: df=87 ss=0.986 a. Should the null hypothesis that all the colors have the same mean weight be​ rejected? b. Does the company have a problem requiring corrective​ action?

a. Yes​, because the​ P-value is less than the significance level. 0.0005<0.05 b. Yes​, because it is likely that the candies do not have equal mean weights.

A sample of colored candies was obtained to determine the weights of different colors. The ANOVA table is shown below. It is known that the population distributions are approximately normal and the variances do not differ greatly. Use a 0.05 significance level to test the claim that the mean weight of different colored candies is the same. If the candy maker wants the different color populations to have the same mean​ weight, do these results suggest that the company has a problem requiring corrective​ action? Source: ​Treatment: df=9 SS=0.135 MS=0.015 Test stat F=2.4436 Critical F=1.9844 pvalue=0.0154 Error: df=91 ss=0.546 ms=0.006 total: df=100 ss=0.681 a.Should the null hypothesis that all the colors have the same mean weight be​ rejected? b. Does the company have a problem requiring corrective​ action?

a.Yes​, because the​P-value is less than the significance level. 0.0154<0.05 b. Yes​, because it is likely that the candies do not have equal mean weights.

Determine whether the following statement is true or false. Sample evidence can prove that a null hypothesis is true.

false: never 100% certain

Determine if the following statement is true or false. When testing a hypothesis using the​ P-value Approach, if the​ P-value is​ large, reject the null hypothesis.

the statement is false

Compute the critical value zα/2 that corresponds to a 96​% level of confidence.

zα/2 =2.05 α=1-.96=.04 zα/2= 4/2=2=0.02 1-.02=.98 invnorm (.98)=2.05

p values are found df=n-1

μ<left:tcf(-99,#,df) μ>Right:tcf(#,99,df) 2tailed:2tcdf(-99,-#,df)