Statistics Test 4

A stock analyst wondered whether the mean rate of return of​ financial, energy, and utility stocks differed over the past 5 years. He obtained a simple random sample of eight companies from each of the three sectors and obtained the​ 5-year rates of return shown in the accompanying table​ (in percent). (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) F0 = ??? (d) Since the​ P-value is ??? there ??? enough evidence to reject the null hypothesis.​ Thus, we ??? conclude that the mean rates of return are different at the α=0.05 level of significance. (e) Pick graph

(a) H0: μfinancial = μenergy = μutilities versus H1: at least one mean is different note: "differed over" = something is different/changed (b) Yes, because there are k=3 simple random​ samples, one from each of k​ populations, the k samples are independent of each​ other, and the populations are normally distributed and have the same variance. note: SAME variance (c) F0 = 2.02 plug values into L1, L2, L3 STAT --> TESTS --> A:ANOVA --> ANOVA(L1, L2, L3) (d) 0.157, is not, cannot (e) Graph fits each data 2nd --> STAT PLOT --> set each to its own L --> Graph

A simple random sample of size n=200 drivers were asked if they drive a car manufactured in a certain country. Of the 200 drivers​ surveyed, 105 responded that they did. Determine if more than half of all drivers drive a car made in this country at the α=0.05 level of significance. (a) State the null and alternative hypotheses. (b) P-value ??? (c) State the conclusion for the test (d) There ??? sufficient evidence at the α=0.05 level of significance to conclude that more than half of all drivers drive a car made in this country.

(a) H0​: p = 0.5 versus H1​: p > 0.5 "more than half" = proportions (p) (b) P-value = .24 STAT --> TESTS --> 5: 1-Prop --> plug in (c) Do not reject H0 because the​P-value is greater than the α=0.05 level of significance. (d) is not do not reject --> is not enough reject --> is enough

According to the Centers for Disease Control and​ Prevention, 9.7​% of high school students currently use electronic cigarettes. A high school counselor is concerned the use of​ e-cigs at her school is higher. ​(a) Determine the null and alternative hypotheses. ​(b) If the sample data indicate that the null hypothesis should not be​ rejected, state the conclusion of the high school counselor. (c) ​Suppose, in​ fact, that the proportion of students at the​ counselor's high school who use electronic cigarettes is 0.266. Was a type I or type II error​ committed?

(a) Hypotheses: H0: p = .097 (change from % to decimal!!!) H1: p > .097 (b) There is not sufficient evidence to conclude that the proportion of high school students exceeds 0.097 at this​ counselor's high school. note: When the null hypothesis is not​ rejected, the alternative hypothesis is not supported (c) A Type II error was committed because the sample evidence led the counselor to conclude the proportion of​ e-cig users was 0.097​, when, in​ fact, the proportion is higher.

According to a food​ website, the mean consumption of popcorn annually by Americans is 56 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 811 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 56 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? Select the correct choice below and fill in the answer box within your choice.

(a) Hypotheses: H0: μ = 56 H1: μ > 56 (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 . 01

Three years​ ago, the mean price of an existing​ single-family home was ​$243,765. A real estate broker believes that existing home prices in her neighborhood are lower. ​(a) Determine the null and alternative hypotheses. ​(b) Explain what it would mean to make a Type I error. ​(c) Explain what it would mean to make a Type II error.

(a) The hypothesis: H0: μ = $243,765 H1: μ < $243,765 (b) The broker rejects the hypothesis that the mean price is $243,765​, when it is the true mean cost. (c) The broker fails to reject the hypothesis that the mean price is ​$243,765​, when the true mean price is less than ​$243,765

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 hypothesis (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) F0 = ??? (d) Since the​ P-value is ??? there is ??? evidence to reject the null hypothesis.​ Thus, we ??? conclude that the means are different at the α=0.01 level of significance.

(a) .H0: μCars=μVans=μSUVs versus H1: at least one mean is different (b) Yes, all the requirements for use of a​ one-way ANOVA procedure are satisfied. (c) F0 = 0.4 plug values into L1, L2, L3 STAT --> TESTS --> A:ANOVA --> ANOVA(L1, L2, L3) (d) 0.6762, insufficient, cannot (e) Graph fits each data 2nd --> STAT PLOT --> set each to its own L --> Graph FIX WINDOW SETTINGS

Twenty years​ ago, 56​% 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 354 of 900 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=0.05 level of significance. (a) Because np0(1−p0)= ??? ??? 10 and the sample size is ??? ​5% of the population​ size, and the sample ???, the requirements for testing the hypothesis ??? satisfied. (b) What are the null and alternative​ hypotheses? (c) z0 = ??? (d) Determine the critical​ value(s). Select the correct choice below and fill in the answer box to complete your choice. (e) Choose the correct conclusion below.

(a) 221.8, >, less than, can be assumed to be, are (b) The hypothesis: H0: p = .56 H1: p ≠ .56 (c) -10.07 1. STAT --> TESTS --> 5:1 Prop-ZTest 2. plug in values (ie. p0 = .56, x = 354, n = 900) (d) ±zα/2 = ±1.96 note: reference normal distribution page level of significance: left-tailed, right-tailed, two-tailed 0.10: -1.28, 1.28, ±1.645 0.05: -1.645, 1.645, ±1.96 0.01: -2.33, 2.33, ±2.575 (e) Reject the null hypothesis. There is sufficient evidence to conclude that the number of parents who feel that students are not being taught enough math and science is significantly different from 20 years ago.

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 0.39. Suppose a random sample of 110 traffic fatalities in a certain region results in 56 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=0.05 level of​ significance? (a) Because np0(1−p0) = ??? >​ 10, the sample size is ??? 5% of the population​ size, and the sample ???, the requirements for testing the hypothesis ??? satisfied (b) Hypothesis? (c) z0 = ??? (d) p-value = ??? (e) Determine the conclusion for this hypothesis test. Choose the correct answer below.

(a) 26.2, less than, is given to be random (go back to see!), are note: 110(0.39)(1 - 0.39) = 26.2 (b) The hypothesis: H0: p = .39 H1: p > .39 (c) z0 = 2.56 1. STAT --> TESTS --> 5:1 Prop-ZTest 2. plug in values (ie. p0 = .39, x = 56, n = 110) (d) p-value = .005 (e) 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

In a previous​ poll, 40​% 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, 454 of 1178 adults with children under the age of 18 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=0.05 significance level. (a) Because np0(1−p0)= ??? ??? 10 and the sample size is ??? ​5% of the population​ size, and the sample ???, the requirements for testing the hypothesis ??? satisfied. (b) What are the null and alternative​ hypotheses? (c) z0 = ??? (d) p-value = ??? (e) 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?

(a) 282.7, >, less than, can be reasonably assumed to be random, are (b) The hypothesis: H0: p = .40 H1: p < .40 (c) z0 = -1.02 1. STAT --> TESTS --> 5:1 Prop-ZTest 2. plug in values (ie. p0 = .40, x = 454, n = 1178) (d) p-value = .153 (e) No​, there is not sufficient evidence because the​ P-value is greater than the level of significance.​ Therefore, do 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 108.6​, and the sample standard deviation is found to be 23.3. Is the population mean greater than 100 at the α=0.025 level of​ significance? (a) Determine the null and alternative hypotheses. (b) Compute the test statistic (c) Determine the P-value (d) ??? the null hypothesis because the​P-value is ??? the level of significance. At the α=0.025 level of​ significance, the population mean ??? ???

(a) H0​: μ = 100 versus H1​: μ > 100 (b) t0 = 2.33 STAT --> TESTS --> 2: T-Test --> plug in (c) P-value = .012 STAT --> TESTS --> 2: T-Test --> plug in (d) reject, less than, is, greater than 100

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=30.7 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) State the null and alternative hypotheses. (b) P-value ??? (c) State the conclusion of the test (d) There ??? sufficient evidence at the α=0.01 level of significance to conclude that the population mean is different from 25.

(a) H0​: μ = 25 versus H1​: μ ≠ 25 note: "is different" --> ≠ (b) P-value = .004 STAT --> TESTS --> 2: T-Test --> plug in (c) Reject H0 because the​ P-value is less than the α=0.01 level of significance. (.004 < .01) (d) is do not reject --> is not enough reject --> is enough

A college entrance exam company determined that a score of 25 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 200 students who completed this core set of courses results in a mean math score of 25.7 on the college entrance exam with a standard deviation of 3.3. Do these results suggest that students who complete the core curriculum are ready for​ college-level mathematics? That​ is, are they scoring above 25 on the math portion of the​ exam? (a) State the appropriate null and alternative hypotheses (b) Verify that the requirements to perform the test using the​ t-distribution are satisfied. (c) Use the classical approach at the α=0.05 level of significance to find the critical value and test the hypotheses. Identify the test statistic. (d) Identify the critical value. (e) Write a conclusion based on the results. ??? the null hypothesis and claim that there ??? sufficient evidence to conclude that the population mean is ??? 25.

(a) H0​: μ = 25versus H1​: μ > 25 (b) The sample size is larger than 30, The​ students' test scores were independent of one another, The students were randomly sampled. (c) t0 = 3 STAT --> TESTS --> 2:T-Test --> plug in (d) ta = 1.66 Use given chart with solve t0 and degrees freedom (149) (do not double because it is only right-tailed) (e) reject, is, greater

One​ year, the mean age of an inmate on death row was 40.9 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 40.3​, with a standard deviation of 8.8. Construct a​ 95% confidence interval about the mean age. What does the interval​ imply? (a) State the null and alternative hypotheses. (b) With​ 95% confidence, the mean age of a death row inmate is between ??? years and ??? years. (c) What does the interval imply?

(a) H0​: μ = 40.9 versus H1​: μ ≠ 40.9 (b) 37.13 and 43.47 STAT --> TESTS --> 8 --> plug in data --> (upper, lower) (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.

A certain vehicle emission inspection station advertises that the wait time for customers is less than 7 minutes. A local resident wants to test this claim and collects a random sample of 49 wait times for customers at the testing station. He finds that the sample mean is 6.58 minutes, with a standard deviation of 3.1 minutes. Does the sample evidence support the inspection​ station's claim? Use the α=0.10 level of significance to test the advertised claim that the wait time is less than 7 minutes. (a) Determine the null and alternative hypotheses. (b) t0 and p-value (c) Since the​ P-value is ??? than α​, ??? the null hypothesis. There ???sufficient evidence to conclude that the mean wait time is less than 7 minutes. In other​ words, the evidence ??? the advertised claim.

(a) H0​: μ = 7 versus H1​: μ < 7 (b) t0 = -.95 and p-value = .174 (c) greater, do not reject, is not, does not support

A credit score is used by credit agencies​ (such as mortgage companies and​ banks) to assess the creditworthiness of individuals. Values range from 300 to​ 850, with a credit score over 700 considered to be a quality credit risk. According to a​ survey, the mean credit score is 704.7. A credit analyst wondered whether​ high-income individuals​ (incomes in excess of​ $100,000 per​ year) had higher credit scores. He obtained a random sample of 37 ​high-income individuals and found the sample mean credit score to be 715.2 with a standard deviation of 82.7. Conduct the appropriate test to determine if​ high-income individuals have higher credit scores at the α=0.05 level of significance. (a) State the null and alternative hypotheses. (b) t0 = ??? (c) p-value = ??? (d) ??? the null hypothesis. There ??? sufficient evidence to claim that the mean credit score of ​high-income individuals is ??? ???

(a) H0​: μ = 704.7 versus H1​: μ > 704.7 (b) t0 = .77 STAT --> TESTS --> 2:T-Test --> plug in (c) p-vale = .222 STAT --> TESTS --> 2:T-Test --> plug in (d) fail to reject, is not, greater than, 704.7

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 50 healthy adults 1 to 4 times daily for 3​ days, obtaining 225 measurements. The sample data resulted in a sample mean of 98.3°F and a sample standard deviation of 0.9°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 null and alternative hypotheses. (b) t0 and p-value? (c) Analyze

(a) H0​: μ = 98.6 versus H1​: μ < 98.6 (b) t0 = -5 and p-value = 0 note: MAKE SURE TO CHECK IF IT IS IN SCIENTFIC NOTATION THEN IT EQUALS 0 (c) Reject H0 since the​ P-value is less than the significance level.

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 64.3 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 0.19. Explain what this value represents. ​(c) Write a conclusion for this hypothesis test assuming an α=0.05 level of significance.

(a) H0​: μ=63.7in. versus H1​: μ>63.7in. (b) There is a 0.19 probability of obtaining a sample mean height of 64.3 inches or taller from a population whose mean height is 63.7 inches. (c) Do not reject the null hypothesis. There is not sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today.

A manufacturer of alloy steel beams requires that the standard deviation of yield strength not exceed 7000 psi. The​ quality-control manager selected a sample of 20 beams and measured their yield strength. The standard deviation of the sample was 8300 psi. Assume that the yield strengths are normally distributed. Does the evidence suggest that the standard deviation of yield strength exceeds 7000 psi at the α=0.01 level of​ significance? (a) Determine the null and alternative hypotheses. (b) (c) (d) There ??? sufficient evidence at the α=0.01 level of significance to conclude that the standard deviation of yield strength exceeds 7000 psi.

(a) H0​: σ = 7000 versus H1​: σ > 7000 (b) 0.112 https://www.socscistatistics.com/pvalues/chidistribution.aspx Determine the value of χ2​, rounding to two decimal places. χ2 = [(n−1)s^2] / σ^2 [(20 - 1)(7000^2)] / 8300^2 = 26.71 2nd VARS --> 6: tcdf --> lower: 26.71 and df: 19 (c) Do not reject H0 because the​ P-value is greater than the α=0.01 level of significance. (d) is not

To test H0​:μ=34 versus H1​:μ≠34​, a simple random sample of size n=40 is obtained. (a) Does the population have to be normally distributed to test this hypothesis by using​ t-distribution methods?​ Why? (b) If x=37.5 and s=9.0​, compute the test statistic (c) Choose correct graph... (d) Approximate and interpret the​ P-value. The probability of observing a ??? statistic ??? the one​ observed, assuming ??? is​ true, is in the range ??? (e) If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, will the researcher reject the null​ hypothesis? Why? Because the​ P-value is ??? α​, the researcher will ??? the null hypothesis. (f) Construct a​ 99% confidence interval to test the hypothesis (g) Because the value ??? lies ??? the confidence​ interval, we ??? the null hypothesis

(a) No—since the sample size is at least​ 30, the underlying population does not need to be normally distributed. (b) 2.46 t0 = (x − μ0) / (s/√n) --> (37.5 - 34) / (9/√40) = 2.46 STAT --> TESTS --> 2 --> plug in values (c) Description: the left and right edges are shaded, not the middle (d) sample, as extreme or more extreme than, H0, 0.01 < P-value < 0.02 note: use 2.46 and df = 40 - 1 = 39, find that 2.46 is between .005 and .01 and double that (bc two-tailed) (e) greater than, fail to reject (f) 33.647, 41.353 STAT --> TESTS --> 8 --> plug in --> (lower, upper) Lower​ bound: x − t(α/2) • s / √n Upper​ bound: x + t(α/2) • s / √n (g) 34, within, fail to reject

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 0.396. The president of a certain college believes that the proportion of students who enroll in her institution have a higher completion rate. (a) State the hypotheses. (b) Which of the following is a Type I Error? (c) Which of the following is a Type II Error?

(a) The hypothesis: H0: p = .396 H1: p > .396​ (b) A Type I error is committed when the null hypothesis is rejected​ when, in​ fact, it is true therefore: The president rejects the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.396​,when, in​ fact, the proportion is 0.396. (c) A Type II error is committed when the null hypothesis is not rejected​ when, in​ fact, the alternative hypothesis is true therefore: The president fails to reject the hypothesis that the proportion of students who earn a​ bachelor's degree within six years is 0.396​, when, in​ fact, the proportion is greater than 0.396.

A can of soda is labeled as containing 16 fluid ounces. The quality control manager wants to verify that the filling machine is not over-filling the cans. Complete parts​ (a) through​ (d) below. ​(a) Determine the null and alternative hypotheses that would be used to determine if the filling machine is calibrated correctly (b) ​The quality control manager obtains a sample of 79 cans and measures the contents. The sample evidence leads the manager to reject the null hypothesis. Write a conclusion for this hypothesis test... There ??? sufficient evidence to conclude that the machine is out of calibration. (c) A ??? has been made since the sample evidence led the​ quality control manager to ??? the null​ hypothesis, when the ??? is true. (d) Management has informed the quality control department that it does not want to shut down the filling machine unless the evidence is overwhelming that the machine is out of calibration. What level of significance would you recommend the quality control manager to​ use? Explain. The level of significance should be ??? because this makes the probability of Type I error ???

(a) The hypothesis: H0: μ = 16 H1: μ > 16 (b) is note: when the null hypothesis is​ rejected, the alternative hypothesis is supported. This means that there is sufficient evidence to conclude that the machine is out of control. (c) Type I error, reject, null hypothesis A Type I error is committed when the null hypothesis is rejected​ when, in​ fact, it is true. A Type II error is committed when the null hypothesis is not rejected​ when, in​ fact, the alternative hypothesis is true. A Type I error has been made since the sample evidence led the​ quality-control manager to reject the null​ hypothesis, when the null hypothesis is true. (d) 0.01 and small

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) Determine the null and alternative hypotheses. ​(b) Explain what it would mean to make a Type I error. ​(c) Explain what it would mean to make a Type II error.

(a) The hypothesis: σ = 1.3 σ < 1.3 (b) The manager rejects the hypothesis that the pressure variability is 1.3​, when it is the true pressure variability. (c) The manager fails to reject the hypothesis that the pressure variability is 1.3​, when the true pressure variability is less than 1.3.

According to a recent article about individuals who have credit​ cards, the mean number of cards per person with credit cards is 4. To test this result a random survey of 60 individuals who have credit cards was conducted. The survey only includes the number of credit cards per participant. The results of the survey are attached below. ​(a) What is the variable of interest in this​ study? Is it qualitative or​ quantitative? The variable of interest is ??? It is a ??? variable. ​(b) Do the results of the survey imply that the mean number of cards per individual is less than 4​? Use the α=0.05 level of significance. (c) t0 and p-value (d) The​ P-value is ??? than the level of significance. ??? the null hypothesis. There ??? sufficient evidence to claim that the mean number of credit cards is ??? 4.

(a) number of credit cards, quantitive (b) H0​: μ = 4 versus H1​: μ < 4 note: < because question says "less than 4" (c) t0 = -0.56 and p-value = .29 plug in values into L1 --> STAT --> TESTS --> 2:T-test --> click DATA --> plug in (d) greater, do not reject, is not, less than

Test the hypothesis using the​ P-value approach. H0: p=0.70 versus H1: p<0.70 n=150, x=95, α=0.01 Perform the test using the​ P-value approach. (a) p-value? (b) What does this mean?

(a) p-value: .0374 1. STAT --> TESTS --> 5:1 Prop-ZTest 2. plug in values (ie. p0 = .7, x = 95, n =150) (b) Since P-value > α, do not reject the null hypothesis note: If P-value < α​, there is sufficient evidence to reject the null hypothesis at the given level of significance. If ​P-value ≥ α​, there is insufficient evidence to reject the null hypothesis at the given level of significance.

To test H0​: μ=20 versus H1​: μ<20​, a simple random sample of size n=18 is obtained from a population that is known to be normally distributed. ​(a) If x = 18.2 and s=4​, compute the test statistic. note: the x has a line above it (b) Draw a​ t-distribution with the area that represents the​ P-value shaded. Which of the following graphs shows the correct shaded​ region? (c) Approximate the P-value (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) t = -1.9 note: t0 = (x − μ0) / (s/√n) --> 18.2 - 20 / (4/√18) = -1.91 (b) Description: normal curve with the area to the left shaded (left-tailed test bc <) (c) 0.025 < p-value < 0.05 1. notice that P(t<-1.9) is the same as P(t>1.9) 2. z-score = 1.9 and degrees of freedom = 1 - 18 = 17 3. that value is between the top row values .025 and .05 (d) The researcher will reject the null hypothesis since the P-value is less than α. note: Reject the null hypothesis only if the​ P-value is less than α. We found that the p-value is between .025 and .05 and, therefore, less than a

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0​: p = 0.3 H1​: p > 0.3 n = 200​, x = 80​, α = 0.05 (a) Calculate the test​ statistic, z0 (b) p-value? (c) Choose the correct result of the hypothesis test for the​ P-value approach below.

(a) z0 = 3.09 note: To test the​ hypothesis, first verify the​ requirements that​ is, the sample must be a simple random​ sample, np0 (1−p0) ≥​ 10, and the sample size cannot be more than​ 5% of the population size. (200)(.3)(1 - .3) = 42 which is more than 10 p^ = x/n --> 80 / 200 = .4 = p^ z0 = p^ - p0 / √[ p0 (1 - p0) / n ] --> 0.4 - 0.3 / √[0.3(1 - 0.3) / 200] = 3.086 (b) P-value = .001 1. STAT --> TESTS --> 5:1 Prop-ZTest 2. plug in values (ie. p0 = .3, x = 80, n =200) (c) Reject the null​hypothesis, because the​P-value is less than α. note: If P-value < α​, there is sufficient evidence to reject the null hypothesis at the given level of significance. If ​P-value ≥ α​, there is insufficient evidence to reject the null hypothesis at the given level of significance.

To test H0: μ=103 versus 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 = 100.0 and s = 5.7​, compute the test statistic. ​(c) Draw a​ t-distribution with the area that represents the​ P-value shaded. (d) Approximate the​ P-value. (e) Interpret the P-value (f) ​If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, will the researcher reject the null​ hypothesis?

(a) ​No, because n ≥ 30 note: Since n=35≥30 and the sample is obtained using simple random​ sampling, the population does not have to be normally distributed to test this hypothesis. (b) -3.11 t0 = (x − μ0) / (s/√n) --> (100 - 103) / (5.7/√35) = -3.11 (c) Description: the left and right edges are shaded, not the middle note: because it is a = vs ≠ it is two tailed which has the two outer parts shaded (d) 0.002<​P-value<0.005 ​P-value=​P(t0<−3.11) + P(t0>3.11​) which is ​2P(t0>3.17​) so since 3.11 is between .0025 and .001 with 34 df, you double that and get .005 and .002 (bc two-tailed) (e) 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. (f) yes note: Reject the null hypothesis only if the​ P-value is less than α. We found that the p-value is between .002 and .005 and, therefore, less than a