Intro to stats chapter 7&8&10
confidence interval (Cl)
- is a range (or an interval) of values used to estimate the true value of a population parameter. aka interval estimate - probability 1 − α
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 300 babies were born, and 255 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
.85 .124 Answer: 0.797 < p < 0.903
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 380 babies were born, and 342 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
.860<p<.940
A data set includes 115 body temperatures of healthy adult humans having a mean of 97.9 degrees * F and a standard deviation of 0.92 degrees * F . Construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of 98.6°F as the mean body temperature?
.99 A= .01 N-1=114 Stat crunch -> calc t-> between-> DF= 114 and set it = to .99 Ta/2= 2.626 2. 2.626• .(.92/square root 115) (.92 is the STD) = .225=E 3. 97.9+-0.225 Answer: 97.675<u<98.125
Find the critical value z α /2 that corresponds to α = 0.04.
0.04 ÷ 2 = 0.02 1.0000 - 0.0200 = 0.9800 Answer: 2.05
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 668 babies were born, and 334 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
0.450<p<.550
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were born, and 340 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
0.85 0.046 Answer: .804<p<.896
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 520 babies were born, and 286 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
1. 286/520= .55 (lower limit) 2. Stat crunch -> normal calc between -> =.99 =2.575 square root 0.55•0.45/520 = 0.056 3) .55-0.056= .55+0.056= Answer: .494 < p< 0.606 No, the pro
A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 435 babies were born, and 348 of them were girls. Use the sample data to construct a 99% confidence interval estimate of the percentage of girls bom Based on the result, does the method appear to be effective?
1. 348/435= .8(lower limit) 2. Stat crunch -> normal calc between -> =.99 =2.575 square root 0.80•0.20/435= 0.049 3) .8-+0.049 .55+0.056= Answer: .751 < p< 0.849 Yes, the proportion of the girls is significantly different from .5
Find the critical value z alpha/2 that corresponds to the given confidence level 83%
1.37
90% critical value
1.645
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t alpha/2 . (b) find the critical value z alpha/2 or ) that neither the normal distribution nor the t distribution applies The confidence level is 99%, o is not known, and the histogram of 63 player salaries in thousands of dollars) of football players on a team is as shown .
1.67
95% critical value
1.96
Find the critical value z Subscript alpha divided by 2 that corresponds to 95%
1.96
Find the critical value z α /2 that corresponds to a 98% confidence level.
2% ÷ 2 = 1% 1.0000 - 0.0100 = 0.9900 Answer: 2.33
99% critical value
2.575
If the confidence level is 92%, what is the value of
8% ÷ 2 = 4% 1.0000 - 0.0400 = 0.9600 z α /2 = 1.75 Answer: 1.75
sample proportion
<p is the best point estimate of the population proportion p.
A magazine provided results from a poll of 1500 adults who were asked to identify their favorite pie. Among the 1500 respondents, 12% chose chocolate pie, and the margin of error was given as plus or minus5 percentage points. Given specific sample data, which confidence interval is wider: the 95% confidence interval or the 80% confidence interval? Why is it wider?
A 95% confidence interval must be wider than an 80% confidence interval in order to be more confident that it captures the true value of the population proportion.
Which of the following is NOT an observation about critical values?
A critical value is the area in the right-tail region of the standard normal curve.
A programmer plans to develop a new software system . In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system . How many computers must be surveyed in order to be 99% confident that his estimate is in error by no more than three percentage points? Complete parts (a) through (c) below
A) 1842 b) Assume that a recent survey suggests that about 91% of computers use a new operating system. P=.91 (2.576)^2(.91)(1-.91)/.03^2 604 Yes, using the additional survey information from part (b) dramatically reduces the sample size.
In a study of the accuracy of fast food drive-through orders, Restaurant Ahad 283 accurate orders and 68 that were not accurate. a. Construct a 95% confidence interval estimate of the percentage of orders that are not accurate. b. Compare the results from part (a) to this 95% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.168 < p < 0.256 . What do you conclude?
A) 283+68=351. 68/351= .1937(lower limit) 2. Stat crunch -> normal calc between -> =.95 =1.96square root 0.1937•0.8063/351= 0.0413 3) .1937-+0.0413= Answer: .152 < p< 0.235 Since the two confidence intervals overlap, neither restaurant appears to have a significantly different percentage of orders that are not accurate.
You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 99% confident that the sample percentage is within 6.5 percentage points of the true population percentage. Complete parts (a) and (b) below.
A) 393 B)Assume that a prior survey suggests that about 29% of air passengers prefer an aisle seat. 324 (2.576)^2(0.29)(1-0.29)/0.065^2
A genetic experiment with peas resulted in one sample of offspring that consisted of 408 green peas and 151 yellow peas. a. Construct a 90% confidence interval to estimate of the percentage of yellow peas. . B. It was expected that 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas is not 25%do the results contradict expectations?
A) 408+151=559. 151/559= .2701(lower limit) 2. Stat crunch -> normal calc between -> =.90 =1.644square root 0.2701•0.7299/559= 0.0309 3) .2701-+0.0309= Answer: .239 < p< 0.301 No, the confidence interval includes 0.25 , so the true percentage could easily equal 25%
Workers at a certain soda drink factory collected data on the volumes in ounces) of a simple random sample of 23 cans of the soda drink. Those volumes have a mean of 12.19 oz and a standard deviation of 0.12 oz, and they appear to be from a normally distributed population. If the workers want the filling process to work so that almost all cans have volumes between 11.89 oz and 12.53 oz, the range rule of thumb can be used to estimate that the standard deviation should be less than 0.16 oz. Use the sample data to test the claim that the population of volumes has a standard deviation less than 0.16 oz. Use a 0.025 significance level. Complete parts (a) through (d) below .
A) ho : o=0.16 mg h1: o <0.16 mg B) x^2= 9.878 Stat-> sample variance -> one sample-> summary Sample variance: 0.22^2 Sample size: 26 Ho: 0.35^2 H1: can not equal Compute Chi-square : 9.877551 C) find P value: 0.006 D) ____h0. There___ sufficient evidence to conclude that the tar content of filtered ——— Reject, is
A simple random sample of 26 filtered 100-mm cigarettes is obtained from a normally distributed population, and the tar content of each cigarette is measured. The sample has a standard deviation of 0.22 mgUse a 0.05 significance level to test the claim that the tar content of filtered 100-mm cigarettes has a standard deviation different from 0.35 mg, which is the standard deviation for unfiltered king- size cigarettes . Complete parts (a) through (d) below.
A) ho : o=0.35 mg h1: o can not = 0.35 mg B) x^2= 9.878 Stat-> sample variance -> one sample-> summary Sample variance: 0.22^2 Sample size: 26 Ho: 0.35^2 H1: can not equal Compute Chi-square : 9.877551 C) find P value: 0.006 D) ____h0. There___ sufficient evidence to conclude that the tar content of filtered ——— Reject, is
Claim: The standard deviation of pulse rates of adult males is more than 11 bpm. For a random sample of 149 adult males, the pulse rates have a standard deviation of 11.6 bpm. Complete parts (a) and (b) below
A) o>11 (bc it says more than) B) Ho: o=11 H2: o>11
Assume a significance level of a= 0.1 and use the given information to complete parts (a) and (b) below. Original claim: The standard deviation of pulse rates of a certain group of adult males IS 74 bpm. The hypothesis test results in a P-value of 0.0034.
A) reject H0 because the P value is less than or equal to a B) there is sufficient evidence to warrant rejection of the claim that the men pulse rate of the group of adult males is 74bpm
Find the critical value z α /2 that corresponds to a 90% confidence level.
Answer: 1.645
Find the critical value z alpha/2 that corresponds to the given confidence level 99%
Answer: 2.58
ClaimThe mean pulse rate in beats per minute) of adult males is equal to 68.6 bmp. For a random sample of 173 adult males, the mean pulse rate is 68.1 bpm and the standard deviation is 10.6 bpm. Complete parts (a) and (b) below.
Deals w population MEAN so use u A) U = 68.6 bmp B) Ho: u= 68.6bpm H1: u can not = 68.6bpm
Claim: The mean pulse rate in beats per minute) of adult males equal to bpmFor random sample of 158 adult males, the mean pulse rate is 67.9 bpm and the standard deviation is 11.3 bpm. Complete parts (a) and (b) below.
Deals w population MEAN so use u A) U = 69 bmp B) Ho: u= 69bpm H1: u can not = 69bpm
Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69.8 bpm. For a random sample of 181 adult males, the mean pulse rate is 71.6 bpm and the standard deviation is 12.3 bpm. Complete parts a) and (b) below. a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information. b. Identify the null and alternative hypotheses.
Deals w population MEAN so use u A) U = 69.8 bmp B) Ho: u= 69.8bpm H1: u can not = 69.8bpm
Assume a significance level of alpha equals 0.1 and use the given information to complete parts (a) and (b) below. Original claim: The standard deviation of pulse rates of a certain group of adult males is more than 18 bpm. The hypothesis test results in a P-value of 0.2551.
Fail to reject Upper H 0 because the P-value is greater than alpha. There is not sufficient evidence to support the claim that the standard deviation of pulse rates of the group of adult males is more than 18 bpm.
An IQ test is designed so that the mean is 100 and the standard deviation is 12 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 90% confidence that the sample mean is within 33 IQ points of the true mean. Assume that σ=12 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
N = .357
The _________ hypothesis is a statement that the value of a population parameter is equal to some claimed value.
Null
Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 824 randomly selected adults showed that 54% of them would erase all of their personal information online if they could . Find the value of the test statistic .
Parameter in this situation is a proportion ^ P = 0.54, p=0.5, q= 1-0.5 -> 0.5, n=824 0.54-0.5 ——————— Square root of (0.5)(0.5)/825 Answer: z= 2.30
magazine provided results from a poll of 1500 adults who were asked to identify their favorite pie. Among the 1500 respondents , 13% chose chocolate pie , and the margin of error was given as plus/minus 5 percentage points . What values do p, a, n, E , and p represent? If the confidence level is 99%, what is the value of a?
Part 1: The value of p is : the sample proportion. The value of q is: found from evaluating 1 - p . The value of n is: The sample size. The value of E is the margin of error. The value of p is the population proportion. Part 2: If the confidence level is 99%, what is the value of a? a: 0.01 OR If the confidence level is 95% what is the value of a? a:0.05
Which of the following groups has terms that can be used interchangeably with the others?
Percentage, Probability, and Proportion
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t infty/2 . (b) find the critical value Za2 , or (c) state that neither the normal distribution nor the distribution applies. The confidence level is 95%, o is not known, and the histogram of 58 player salaries in thousands of dollars) of football players on a team is as shown.
Stat crunch-> calculator t-> between-> plug in 1-58 in DF & .95 after = Answer: 2
Claim: The standard deviation of pulse rates of adult males is less than 13 bpm. For a random sample of 192 adult males, the pulse rates have a standard deviation of 11.6 bpm. Complete parts (a) and (b) below
The claim involves population standard deviation so use o A) o< 13 bpm B) Ho: o=13 H2: o<13
A newspaper provided a "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustration showed that 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company. The margin of error was given as plus or minus 3 percentage points. What important feature of the poll was omitted?
The confidence level
Which of the following is NOT true about the tails in a distribution?
The inequality symbol in the alternative hypothesis points away from the critical region.
Which of the following is NOT a property of the chi-square distribution?
The meaning of a chai Square distribution zero
Claim: The MEAN pulse rate (in beats per minute) of adult males is equal to 69 bpm. For a random sample of 181 adult males, the mean pulse rate is 70.6 bpm and the standard deviation is 12.3 bpm. Find the value of the test statistic.
The parameter in this situation is a mean that follows the t-distribution - X = 70.6, u= 69, s= 12.3, n=181 70.6-69 ————- 12.3 divided by square root of 181 Answer: 1.75
ClaimFewer than 89% of adults have a cell phone. In a reputable poll of 1250 adults, 85% said that they have a cell phone. Find value of the test statistic.
The perimeter in this situation is a proportion ^ P = 0.85, p=0.89, q= 1-0.89-> 0.11, n=1250 0.85-0.89 —————— Square root of (0.89)(0.11) ——————- 1250 Answer: -4.52
Claim: The standard deviation pulse rates of males less than 13 bpmFor a random sample of 192 adult males, the pulse rates have a standard deviation of 11.6 bpmFind the value of the test statistic.
The perimeter of the situation is in standard deviation N=192, s= 11.6, o= 13 (192-1)(11.6)^2 X^2= ———— 13^2 Answer:152.08
Which of the following is NOT a requirement for testing a claim about a standard deviation or variance?
The population must be skewed to the right
A magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 respondents, 13% chose chocolate pie, and the margin of error was given as plus or minus5 percentage points. Describe what is meant by the statement that "the margin of error was given as plus or minus5 percentage points."
The statement indicates that the interval 13%plus or minus5% is likely to contain the true population percentage of people that prefer chocolate pie.
Which of the following is NOT a requirement for constructing a confidence interval for estimating the population proportion?
The trials are done without replacement.
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t alpha/2 . (b) find the critical value z alpha/2 or ) that neither the normal distribution nor the t distribution applies The confidence level is 99%, o is not known, and the histogram of 60 player salaries in thousands of dollars) of football players on a team is as shown .
To be a t instead of z, the value has to be greater than 30 & 60 is greater than 30 & the STD must be unknown Ta/2= 2.66
Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value t alpha/2 . (b) find the critical value z alpha/2 or ) that neither the normal distribution nor the t distribution applies The confidence level is 99%, o is not known, and the histogram of 81 player salaries in thousands of dollars) of football players on a team is as shown .
To be a t instead of z, the value has to be greater than 30 & 81 is greater than 30 & the STD must be unknown Ta/2= 2.64
Identify the type 1 error and the type 2 error that corresponds to the given hypothesis. The proportion of people who write with their left hand is equal to 0.15.
Type 1 error: Reject the claim that the proportion of people who write with their left hand is 0.15 when the proportion is actually 0.15. Type 2 error: Fail to reject the claim that the proportion of people who write with their left hand is 0.15 when the proportion is actually different from 0.15.
Identify the type 1 error and the type 2 error that correspond to the given hypothesis. The percentage of high school students who graduate is less than 55 %.
Type 1 error: Reject the null hypothesis that the percentage of high school students who graduate is equal to 55 % when that percentage is actually equal to 55 %. Type 2 error: Fail to reject the null hypothesis that the percentage of high school students who graduate is equal to 55 % when that percentage is actually less than 55 %.
Identify the type 1 error and the type 2 error that corresponds to the given hypothesis. The proportion of people who have a job is less than 88%
Type 1 error: Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually EQUAL to 88 % Type 2 error: Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88% when that percentage is actually LESS THAN 88%.
Identify the type 1 error and the type 2 error that correspond to the given hypothesis. The percentage of adults who retire at age 65 is equal to 62 %. Type 1 error: Reject the null hypothesis that the percentage of adults who retire at age 65 is equal to 62 % when that percentage is actually equal to 62 %
Type 2 error: Fail to reject the null hypothesis that the percentage of adults who retire at age 65 is equal to 62 % when that percentage is actually different from 62 %.
We must be careful to interpret confidence intervals correctly. There is a correct interpretation and many different and creative incorrect interpretations of the confidence interval 0.405 < p < 0.455
We are 95% confident that the interval from 0.405 to 0.455 actually does contain the true value of the population proportion p
The piston diameter of a certain hand pump is 0.6 inch. The manager determines that the diameters are normally distributed, with a mean of 0.6 inch and a standard deviation of 0.006 inch. After recalibrating the production machine, the manager randomly selects 23 pistons and determines that the standard deviation is 0.0041 inch. Is there significant evidence for the manager to conclude that the standard deviation has decreased at the alpha = 0.05 level of significance?
What are the correct hypotheses for this test? The null hypothesis is H 0 : sigma = 0.006 The alternative hypothesis is H 1 < 0.006 Calculate the value of the test statistic . X^2:
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.5 degrees * F and a standard deviation of 0.61 ^ 0 * F Construct a 99% confidence interval estimate of the mean body temperature of all healthy humansWhat does the sample suggest about the use of 98.6 degrees * F as the mean body temperature?
What is the confidence interval estimate of the population mean u? __^ F< u< ___^ F .99 N-1=102 Stat crunch -> calc t-> between-> DF= 102 and set it = to .99 Ta/2= 2.6249 2. 2.6204• .(.61/square root 103) = .0155 3. 2.6249+-.0155 Answer: 98.342<u<98.658 This is just such a mean body temperature could very possibly by 98.6
A 9-year-old girl did a science fair experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapists to identify the selected hand by placing their hand just under her hand without seeing it and without touching it. Among 261 trials, the touch therapists were correct 109 times. Use a 0.05 significance level to test the claim that touch therapists use a method EQUIVALENT to random guesses. Do the results suggest that touch therapists are effective?
Which of the following is the hypothesis test to be conducted? H0: p= 0.5 H1: p can not =0.5 Identify the test statistic for the hypothesis test
In a study of 809 randomly selected medical malpractice lawsuits, it was found that 472 of them were dropped or dismissedUse a 0.01 significance level to test the claim that most medical malpractice lawsults dropped of dismissed.
Which of the following is the hypothesis test to be conducted? H0: p= 0.5 H1: p>0.5 What is the test statistic?
Find the critical value z alpha/2 that corresponds to the given confidence level 94%
You have an alpha level of 0.06a/2 = 0.06/2 = 0.030.05 - 0.03 = 0.4700 now look up the z value for p=0.4700 in the z-table and thats your answer. ANSWER: 1.88
We can use a sample proportion to
a confidence interval to estimate the true value of a population proportion, and we should know how to interpret such confidence intervals.
A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 90% confident that his estimate is in error by no more than three percentage points question mark Complete parts (a) through (c) below.
a) Assume that nothing is known about the percentage of computers with new operating systems. n=752 b) Assume that a recent survey suggests that about 90% of computers use a new operating system. n=271 c) Does the additional survey information from part (b) have much of an effect on the sample size that is required? Yes, using the additional survey information from part (b) dramatically reduces the sample size.
A drug is used to help prevent blood clots in certain patients. In clinical trials, among 4568 patients treated with the drug, 138 developed the adverse reaction of nausea. Construct a 95%confidence interval for the proportion of adverse reactions.
a) Find the best point estimate of the population proportion p 138/4568-> 0.030 b) Identify the value of the margin of error Stat crunch -> normal calc between -> =.95 Answer: 0.005 c) Construct the confidence interval. 0.025< p < .035 d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion .
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n = 1013 and x = 532 who said "yes" a 95% confidence
a) Find the best point estimate of the population proportion p Answer: 532/1013 -> .525 b) Identify the value of the margin of error Stat crunch -> normal calc between -> =.95-> 1.96 Answer: .031 c) Construct the confidence interval. 0.494 < p < .556 d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion .
In a science fair project, Emily conducted an experiment in which she tested professional touch therapists to see if they could sense her energy field. She flipped a coin to select either her right hand or her left hand, and then she asked the therapists to identify the selected hand by placing their hand just under Emily's hand without seeing it and without touching it. Among 350 trials, the touch therapists were correct 169 times. Complete parts (a) through (d).
a) Given that Emily used a coin toss to select either her right hand or her left hand, what proportion of correct responses would be expected if the touch therapists made random guesses? _____ .5 b) Using Emily's sample results, what is the best point estimate of the therapists' success rate? _____ 169/350 = .483 c) Using Emily's sample results, construct a 95% confidence interval estimate of the proportion of correct responses made by touch therapists. _____ < p < _____ Stat crunch -> normal calc between -> =.95 =1.96square root 0.483•0.517/350= 0.0524 3) .483-+0.0524= .431<p<.535 What do the results suggest about the ability of touch therapists to select the correct hand by sensing energy fields? A) Since the confidence interval is not entirely above 0.5, there does not appear to be sufficient evidence that touch therapists can select the correct hand by sensing energy fields.
You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 95% confident that the sample percentage is within 3.5 percentage points of the true population percentage. Complete parts (a) and (b) below.
a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. z=1.96 1.96^2 *.25 / .035^2(percentage point) 784 b. Assume that a prior survey suggests that about 31% of air passengers prefer an aisle seat. change .25 to .31(1-.31) 671
You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 99% confident that the sample percentage is within 4.5 percentage points of the true population percentage. Complete parts (a) and (b) below.
a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. z=2.5756 2.576^2 *.25 / .045^2(percentage point) 819 b. Assume that a prior survey suggests that about 32% of air passengers prefer an aisle seat. (2.576)^2(0.32)(1-0.32)/0.045^2 714
Assume a significance level of alpha = 0.05 and use the given information to complete parts (a) and () below. Original claim: More than 44% of adults would erase all of their personal information online if they could. The hypothesis test results in a P-value 0.1606
a. State a conclusion about the null hypothesis. (Reject Upper H 0 or fail to reject Upper H 0.) Choose the correct answer below. FAIL to reject Upper H 0 because the P-value is greater than alpha. b. Without using technical terms, state a final conclusion that addresses the original claim. Which of the following is the correct conclusion? There is NOT sufficient evidence to support the claim that the percentage of adults that would erase all of their personal information online if they could is less than 44%.
Claim: The standard deviation of pulse rates of adult males is more than 12 bpm. For a random sample of 151 adult males, the pulse rates have a standard deviation of 12.6 bpm. a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses.
a. o>12 bpm b. o=12 o>12
Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 824 randomly selected adults showed that 46% of them would erase all of their personal information online if they could a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information. b. Identify the null and alternative hypotheses.
a. p<0.5 b. h0: p=0.5 h1: p<0.5
Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 557 randomly selected adults showed that 61% of them would erase all of their personal information online if they could a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information. b. Identify the null and alternative hypotheses.
a. p>0.5 b. h0: p=0.5 h1: p> 0.5
Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 629 randomly selected adults showed that 63% of them would erase all of their personal information online if they could a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information. b. Identify the null and alternative hypotheses.
a. p>0.5 b. h0: p=0.5 h1: p> 0.5
Claim: The mean pulse rate (in beats per minute) of adult males is equal to 68.7 bpm. For a random sample of 167 adult males, the mean pulse rate is 68.5 bpm and the standard deviation is 11.2 bpm. Complete parts (a) and (b) below. a. Express the original claim in symbolic form. b. Identify the null and alternative hypotheses.
a. u=68.7 bpm b. h0; u=68.7 h1 does not equal 68.7
Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 504 randomly selected adults showed that 99.7% of them would erase all of their personal information online if they could. Make a subjective estimate to decide whether the results are significantly low or significantly high, then state a conclusion about the original claim.
are, high, is
The value of Modifying Above <q with caret is
found from 1- <p
A _____________ is a procedure for testing a claim about a property of a population.
hypothesis test
margin of error
likely difference (with probability 1 - α, such as 0.95) between the observed proportion the and the true value of the population proportion <p
A _______ is a single value used to approximate a population parameter.
point estimate
The value of p is
population proportion
Assume a significance level of alpha equals 0.1 and use the given information to complete parts (a) and (b) below. Original claim: The mean pulse rate (in beats per minute) of a certain group of adult males is 71 bpm. The hypothesis test results in a P-value of 0.0027.
reject Upper H 0 because the P-value is less than or equal to alpha. There is sufficient evidence to warrant rejection of the claim that the mean pulse rate (in beats per minute) of the group of adult males is 71 bpm.
z x/2
s a critical value that is a z-score that separates an area of x/2 in the right tail of the standard normal distribution
critical value
s a z-score on the borderline separating sample statistics that are unlikely to occur (or significantly high or low) from those that are likely to occur(or not significant)
The ___________ is a value used in making a decision about the null hypothesis and is found by converting the sample statistic to a score with the assumption that the null hypothesis is true.
test statistic
The value of E is
the margin of error
The value of ModifyingAbove <p with caret is
the sample proportion
The value of n is
the sample size
99% (or 0.99) confidence level:
α = 0.01
95% (or 0.95) confidence level:
α = 0.05
90% (or 0.90) confidence level:
α = 0.10
A programmer plans to develop a new software system . In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system . How many computers must be surveyed in order to be 95 % confident that his estimate is in error by no more than three percentage points? Complete parts (a) through (c) below
a) Assume that nothing is known about the percentage of computers with new operating E is 0.03 (1.96)^2(.25)/0.03^2 1068 B) 241
A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 99% confident that his estimate is in error by no more than two percentage points question mark Complete parts (a) through (c) below.
a) Assume that nothing is known about the percentage of computers with new operating systems. n = 4147 b) Assume that a recent survey suggests that about 98% of computers use a new operating system. n = 326 c) Does the additional survey information from part (b) have much of an effect on the sample size that is required? Yes, using the additional survey information from part (b) dramatically reduces the sample size.
