Chapter 9 - Inference From Two Samples

A Hypothesis Test for the difference of two means from dependent samples: Paired T-test

(1) subtract each data pair to create one data set of differences (2) find the mean of these differences, denoted d-bar (3) find the standard deviation of the differences, denoted s_d (4) the standard deviation for the test is, s_d/√n

What StatCrunch computes: CI for p1 - p2

(p1 - p2) +- E E = z_a/2 • std. error for this test

Confidence Level based Hyp.Test's alpha

- If hypothesis test was one-tailed, balance with a matching area in the opposite tail - Use confidence level based on area remaining between balanced tails

A Hypothesis Test for the difference of two means is a Two-Sample T-test

- if the two samples are randomly selected - If the two samples are independent - if the two populations each have a normal distribution or both sample sizes are large (n1 > 30 and n2 > 30)

A Hypothesis Test for the difference of two means is a Paired T-test

- if the two samples are randomly selected - if both populations each are normally distributed or both samples are large (n > 30) - if the two samples are dependent ("paired")

The data below are yields for two different types of corn seed that were used on adjacent plots of land. Assume that the data are simple random samples and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval estimate of the difference between type 1 and type 2 yields. What does the confidence interval suggest about farmer​ Joe's claim that type 1 seed is better than type 2​ seed?

21.53 < u_d < 103.97 Because the confidence interval only includes positive values and does not include ​zero, there is sufficient evidence to support farmer​ Joe's claim.

What does the confidence interval suggest about farmer​ Joe's claim that type 1 seed is better than type 2​ seed? -5.81 < u_d < 73.56

Because the confidence interval includes ​zero, there is not sufficient evidence to support farmer​ Joe's claim.

When the bounds of the Confidence Interval include zero Negative endpoint< p1 - p2 <Positive endpoint ...the computed difference is less than the margin or error:

Because zero is in the interval We say there is NO SIGNIFICANT DIFFERENCE between the two proportions.

To find a confidence interval for iu_d that predicts: How different are the two means?

CI for u_d (the average difference in the population) --by hand: d-bar +- E where E = t_a/2 * (s_d/√n) StatCrunch: STAT > T stats > Paired

Compare average score on standardized test for a group of 50 students before and after a study program.

Dependent

Compare memory test results of treatment group before and after taking herbal supplements.

Dependent

Compare Internet time for male vs. female students.

Independent

Compare mean weight loss of 40 clients from Weight Watchers vs. 32 clients from Jenny Craig.

Independent

Compare memory test results of treatment group taking herbal supplements vs. control group taking placebo.

Independent

Compare the average price of a one night stay at a Holiday Inn Express Hotel and Red Roof Inn Hotel based on samples of 8 of each chosen randomly across the country

Independent

Several students were tested for reaction times​ (in thousandths of a​ second) using their right and left hands.​ (Each value is the elapsed time between the release of a strip of paper and the instant that it is caught by the​ subject.) Results from five of the students are included in the graph to the right. Use a 0.20 significance level to test the claim that there is no difference between the reaction times of the right and left hands

Let u_d be the mean of the differences of the right and left hand reaction times. u_d = 0 u_d /= 0 -7.309 t = +-1.533 (two-tail) There is enough evidence to warrant rejection of the claim that there is no difference between the reaction times of the right and left hands.

Researchers collected data on the numbers of hospital admissions resulting from motor vehicle​ crashes, and results are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Use a 0.05 significance level to test the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

Let u_d be the mean of the differences of the right and left hand reaction times. u_d = 0 u_d /= 0 t = +-2.571 (two-tail) There is sufficient evidence to warrant rejection of the claim of no effect. Hospital admissions appear to be affected.

a. Use a 0.05 significance​ level, and test the claim that the mean longevity for archbishops is less than the mean for monarchs after coronation.

Reject the null hypothesis. There is sufficient evidence to support the claim that archbishops have lower mean longevity than monarchs. [use .9] Yes, because the confidence interval contains only negative values.

When games were sampled throughout a​ season, it was found that the home team won 116 of 173 lacrosse ​games, and the home team won 51 of 76 soccer games. The result from testing the claim of equal proportions are shown on the right. Does there appear to be a significant difference between the proportions of home​ wins? What do you conclude about the home-field​ advantage? Does there appear to be a significant difference between the proportions of home​ wins? (Use the level of significance alpha equals 0.05) What do you conclude about the home-field​ advantage? (Use the level of significance alpha equals 0.05)

Since the​ p-value is large​, there is not a significant difference. The advantage appears to be about the same for lacrosse and soccer.

To find a confidence interval for difference in two population proportions, p1 - p2 How different are the two proportions?

StatCrunch> Proportion Stats > 2 Samples> With Summary use Options: Edit (to re-use info previously entered). ...click: Confidence Interval

In a random sample of​ males, it was found that 25 write with their left hands and 214 do not. In a random sample of​ females, it was found that 71 write with their left hands and 450 do not. Use a 0.01 significance level to test the claim that the rate of​ left-handedness among males is less than that among females. Complete parts​ (a) through​ (c) below.

The​ P-value is greater than the significance level of alpha equals0.01​, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the rate of​ left-handedness among males is less than that among females. Because the confidence interval limits include ​0, it appears that the two rates of​ left-handedness are equal. There is not sufficient evidence to support the claim that the rate of​ left-handedness among males is less than that among females. The rate of​ left-handedness among males does not appear to be less than the rate of​ left-handedness among females.

Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. The second sample consists of 2100 people with 1529 of them having the same common attribute. Compare the results from a hypothesis test of p 1 equals p 2 ​(with a 0.01 significance​ level) and a 99​% confidence interval estimate of p 1 minus p 2.

crit vals: -2.576, 2.576 The test statistic is not in the critical​ region, so fail to reject the null hypothesis. There is insufficient evidence to conclude that p 1 /= p 2. Since 0 is included in the​ interval, it indicates to fail to reject the null hypothesis. The results are the same​, since the hypothesis test suggests that p 1 = p 2​, and the confidence interval suggests that p 1 = p 2.

Two samples are​ ________________ if the sample values are paired.

dependent

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2761 occupants not wearing seat​ belts, 39 were killed. Among 7728 occupants wearing seat​ belts, 11 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 > p2 The​ P-value is less than the significance level of alpha equals0.05​, so reject the null hypothesis. There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts. use 90% confidence interval .009 < (p1-p2) < .016 Because the confidence interval limits do not include ​0, it appears that the two fatality rates are not equal. Because the confidence interval limits include only positive ​values, it appears that the fatality rate is higher for those not wearing seat belts. The results suggest that the use of seat belts is associated with lower fatality rates than not using seat belts.

Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Refer to the accompanying data set. Use a 0.01 significance level to test the claim that the sample of home voltages and the sample of generator voltages are from populations with the same mean. If there is a statistically significant​ difference, does that difference have practical​ significance?

u1 = u2 u1 /= u2 Reject H0. There is sufficient evidence to warrant rejection of the claim that the sample of home voltages and the sample of generator voltages are from populations with the same mean. The difference is statistically significant. The sample means suggest that the difference does not have practical significance. The generator could be used as a substitute when needed.

Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.10 to test for a difference between the number of words spoken in a day by each member of 30 different couples.

u_d = 0 u_d /= 0 2.94, .006 Since the​ P-value is less than the significance​ level, reject the null hypothesis. There is sufficient evidence to support the claim that there is a difference between the number of words spoken in a day by each member of 30 different couples.

A study was conducted to determine the effectiveness of a certain treatment. A group of 112 patients were randomly divided into an experimental group and a control group. The table shows the result for their net improvement.

Yes, because the test statistic is in the critical region 4.89 to 10.71 We are 98​% confident that mu1 is greater than mu2 by between 4.89 and 10.71

The Confidence interval for u1 - u2 that predicts: How much lower is the average test score under Teaching Method A compared to Method B?

Confidence Level alpha =.05 (left-tailed) ->balanced with .05 in right ->Use 90% Confidence Interval to estimate the difference

The Confidence interval for u1 - u2 that predicts: How much higher are the repair costs for the small pickups compared to the SUVs?

Confidence Level alpha =.05 (right-tailed) -> balanced with .05 in left -> Use 90% Confidence Interval to estimate the difference

a. Use a 0.01 significance level to test the claim that the mean amount of​ strontium-90 from city​ #1 residents is greater than the mean amount from city​ #2 residents.

Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean amount of​ strontium-90 from city​ #1 residents is greater. [use .98] Yes, because the confidence interval contains zero.

When games were sampled throughout a​ season, it was found that the home team won 115 of 174 soccer ​games, and the home team won 65 of 77 hockey games. The result from testing the claim of equal proportions are shown on the right. Does there appear to be a significant difference between the proportions of home​ wins? What do you conclude about the home field​ advantage?

p-val < a Since the​ p-value is small​, there is a significant difference. The advantage appears to be higher for hockey.

Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 33 of the 39 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 89 of the 103 subjects developed rhinovirus infections. Use a 0.01 significance level to test the claim that echinacea has an effect on rhinovirus infections. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 /= p2 The​ P-value is greater than the significance level of alpha equals0.01​, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect. Because the confidence interval limits include ​0, there does not appear to be a significant difference between the two proportions. There is not evidence to support the claim that echinacea treatment has an effect. Echinacea does not appear to have a significant effect on the infection rate.

Two different simple random samples are drawn from two different populations. The first sample consists of 40 people with 18 having a common attribute. The second sample consists of 2200 people with 1547 of them having the same common attribute. Compare the results from a hypothesis test of p 1 equals p 2 ​(with a 0.05 significance​ level) and a 95​% confidence interval estimate of p 1 minus p 2.

p1 = p2 p1 /= p2 z = -3.46 crti val: -1.96, 1.96 The test statistic is in the critical​ region, so reject the null hypothesis. There is sufficient evidence to conclude that p 1 /= p 2. -.409 < (p1-p2) -.098 Since 0 is not included in the​ interval, it indicates to reject the null hypothesis. The results are the same​, since the hypothesis test suggests that p 1 /= p 2​, and the confidence interval suggests that p 1 /= p 2.

Since an instant replay system for tennis was introduced at a major​ tournament, men challenged 1401 referee​ calls, with the result that 423 of the calls were overturned. Women challenged 771 referee​ calls, and 220 of the calls were overturned. Use a 0.01 significance level to test the claim that men and women have equal success in challenging calls. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 /= p2 z = .81 p-val = .418 The​ P-value is greater than the significance level of alpha = 0.01​, so fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls. -.036 < p1 - p2 < .069 Because the confidence interval limits include ​0, there does not appear to be a significant difference between the two proportions. It appears that men and women have equal success in challenging calls.

In a randomized controlled​ trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 310 infants using​ bednets, 21 developed malaria. Among 260 infants not using​ bednets, 27 developed malaria. Use a 0.05 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be​ effective? Conduct the hypothesis test by using the results from the given display.

p1 = p2 p1 < p2 test stat: z = -1.55 p-val = 0.061 The​ P-value is greater than the significance level alpha​, so fail to reject the null hypothesis. There is insufficient evidence to support the claim that the incidence of malaria is lower for infants using bednets. The bednets do not appear to be effective.

In a randomized controlled​ trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 327 infants using​ bednets, 12 developed malaria. Among 281 infants not using​ bednets, 26 developed malaria. Use a 0.05 significance level to test the claim that the incidence of malaria is lower for infants using bednets. Do the bednets appear to be​ effective?

p1 = p2 p1 < p2 -2.84 .002 The​ P-value is less than the significance level alpha​, so reject the null hypothesis. There is sufficient evidence to conclude that the incidence of malaria is lower for infants using bednets. The bednets appear to be effective.

Requirements: 2 Proportion Test If independent samples are taken from two populations and the samples are large enough you can test for the difference between population proportions p1 - p2

- if the two samples are randomly selected - if the two samples are independent - if the samples are large enough to use a normal sampling distribution: That is: both samples have at least 5 successes & 5 failures

A study was conducted to measure the effectiveness of hypnotism in reducing pain. The measurements are centimeters on a pain scale before and after hypnosis. Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval for the mean of the ​"before - ​after" differences. Does hypnotism appear to be effective in reducing​ pain?

-.57 < u_d < 2.57 No, because the confidence interval includes zero

Listed below are the heights of candidates who won elections and the heights of the candidates with the next highest number of votes. The data are in chronological​ order, so the corresponding heights from the two lists are matched. Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval estimate of the mean of the population of all​ "winner/runner-up" differences. Does height appear to be an important factor in winning an​ election?

-1.57 < u_d < 1.07 No, because the confidence interval includes zero

Data on the numbers of hospital admissions resulting from motor vehicle crashes are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Construct a​ 95% confidence interval estimate of the mean of the population of differences between hospital admissions. Use the confidence interval to test the claim that when the 13th day of a month falls on a​ Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

-13.31 < u_d < -4.29 Yes​, because the confidence interval does not include zero.

Given in the table are the BMI statistics for random samples of men and women. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.01 significance level for both parts.

Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that men and women have the same mean BMI. [use .99] Yes, because the confidence interval contains zero.

Compare on-task time for students in blue room vs. same class in red room.

Dependent

Compare the average price of a one night stay at a Holiday Inn Express Hotel to a Red Roof Inn Hotel by randomly selecting 8 towns that have both types of hotels, and recording the price for each.

Dependent

An engineer studied the strength of two different mixes of concrete​ (in psi). He randomly sampled the strength of 9 cylinders made with mix 1 and 10 cylinders made with mix 2.

No, because the test statistic is not in the critical region 655 to 1099 We are 90​% confident that the mean strength mix of 2 is greater than the mean strength of mix 1 by between 655 and 1099 psi

dependent

Samples are dependent when individuals in one sample are used to determine the individuals in the second sample. Each data value of one sample is "matched" or "paired" with a data value of the other sample. Usually a pre-existing pairing (couples, twins, siblings) or the same person/same subject measured under different circumstances (before & after, with tutoring & without)

Several students were tested for reaction times​ (in thousandths of a​ second) using their right and left hands.​ (Each value is the elapsed time between the release of a strip of paper and the instant that it is caught by the​ subject.) Results from five of the students are included in the graph to the right. Use a 0.05 significance level to test the claim that there is no difference between the reaction times of the right and left hands.

The critical values are t = +- 2.776 There is not enough evidence to warrant rejection of the claim that there is no difference between the reaction times of the right and left hands.

When each value from one sample is paired with a data value in the second sample

The differences, d = x1 - x2 are calculated for each data pair. This reduces the original data down to one data set of differences (no longer two data sets)

A data set includes the age at marriage for 210 randomly selected married couples

The samples are dependent because there is a natural pairing between the two samples.

Determine whether the samples are independent or dependent. A data set includes the morning and evening temperature for the last 120 days.

The samples are dependent because there is a natural pairing between the two samples.

Determine whether the samples are independent or dependent. To test the effectiveness of a drug comma cholesterol levels are measured in 150 men before and after the treatment.

The samples are dependent because there is a natural pairing between the two samples.

A researcher wishes to compare IQ scores of married doctors and their spouses. She obtains a random sample of 482 such couples who take an IQ test and determines each spouse's IQ.

The sampling is dependent because the individuals selected to be in one sample are used to determine the individuals to be used in the second sample

t = -2.928 critical t-values = +-2.571

This data provides significant evidence to conclude there is a difference in the numbers of hospital admissions based on Friday dates

An educator wants to determine whether a new curriculum significantly improves standardized test scores for third grade students. She randomly divides 90 third dash graders into two groups. Group 1 is taught using the new ​curriculum, while group 2 is taught using the traditional curriculum. At the end of the school​ year, both groups are given the standardized test and the mean scores are compared. Determine whether the sampling is dependent or independent.

This sampling method is independent.

dependent. (matched-pairs)

Two samples are dependent if each member of one sample corresponds to a member of the other sample.

independent.

Two samples are independent if the sample selected from one population is not related to the sample selected from the 2nd population.

independent

Two samples are independent when members of one sample do not dictate which individuals are to be in the second sample.

alpha =.05 (left-tailed) -> balanced with .05 right

Use 90% Confidence Interval to estimate the difference between the proportions

A researcher studies water clarity at the same location in a lake on the same dates during the course of a year and repeats the measurements on the same dates 5 years later. The researcher immerses a weighted disk painted black and white and measures the depth​ (in inches) at which it is no longer visible. The collected data is given in the table below. Complete parts ​(a) and ​(b) below.

Using the same dates makes the second sample dependent on the first. Do not reject H0

When the bounds of the Confidence Interval include zero: Negative endpoint< p1 - p2 <Positive endpoint

We say there is NO SIGNIFICANT DIFFERENCE between the two proportions.

since this interval includes zero...

With 90% confidence, this data predicts NO SIGNIFICANT DIFFERENCE between the percentage of smokers in California and smokers in Oregon.

An engineer studied the strength of two different mixes of concrete​ (in psi). He randomly sampled the strength of 9 cylinders made with mix 1 and 10 cylinders made with mix 2..

Yes​, because the test statistic is in the critical region 645 to 1131 We are 90​% confident that the mean strength mix of 2 is greater than the mean strength of mix 1 by between 645 and 1131 psi p-val = 0

Use the weights of cans of generic soda as sample​ one, and use the weights of cans of the diet version of that soda as sample two. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Construct a 95​% confidence interval estimate of the difference between the mean weight of the cans of generic soda and the mean weight of cans of the diet version of that soda. Does there appear to be a difference between the mean​ weights?

[use .95] The mean weight for the generic soda appears to be greater than the mean weight for the diet variety because the confidence interval contains only positive values.

Two samples are​ ____________ if the sample values from one population are not related to or somehow naturally paired or matched with the sample values from the other population.

independent

In a study of treatments for very painful​ "cluster" headaches, 146 patients were treated with oxygen and 148 other patients were given a placebo consisting of ordinary air. Among the 146 patients in the oxygen treatment​ group, 123 were free from headaches 15 minutes after treatment. Among the 148 patients given the​ placebo, 38 were free from headaches 15 minutes after treatment. Use a 0.05 significance level to test the claim that the oxygen treatment is effective. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 > p2 The​ P-value is less than the significance level of alpha equals0.05​, so reject the null hypothesis. There is sufficient evidence to support the claim that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. Because the confidence interval limits do not include ​0, it appears that the two cure rates are not equal. Because the confidence interval limits include only positive ​values, it appears that the cure rate is higher for the oxygen treatment than for the placebo. The results suggest that the oxygen treatment is effective in curing​ "cluster" headaches.

A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 304 people over the age of​ 55, 72 dream in black and​ white, and among 297 people under the age of​ 25, 13 dream in black and white. Use a 0.05 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 > p2 [in stat-crunch, p1-p2 = 0, p1 - p2 > 0] z = 6.79 p-val = 0 The​ P-value is less than the significance level of alpha = 0.05​, so reject the null hypothesis. There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. .148 < (p1-p2) < .238 Because the confidence interval limits do not include ​0, it appears that the two proportions are not equal. Because the confidence interval limits include only positive ​values, it appears that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25. No. The results speak to a possible difference between the proportions of people over 55 and under 25 who dream in black and​ white, but the results cannot be used to verify the cause of such a difference.

A simple random sample of​ front-seat occupants involved in car crashes is obtained. Among 2994 occupants not wearing seat​ belts, 37 were killed. Among 7677 occupants wearing seat​ belts, 16 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. Complete parts​ (a) through​ (c) below.

p1 = p2 p1 > p2 z = 6.78 p-val = 0 The​ P-value is less than the significance level of alpha equals 0.05​, so reject the null hypothesis. There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts. .006 < p1 - p2 < .014 Because the confidence interval limits do not include ​0, it appears that the two fatality rates are not equal. Because the confidence interval limits include only positive ​values, it appears that the fatality rate is higher for those not wearing seat belts. The results suggest that the use of seat belts is associated with lower fatality rates than not using seat belts.

When the bounds of the Confidence Interval are both positive: positive endpoint < p1 - p2 < positive endpoint

p1 must be greater than p2 by between lower bound and upper bound

When the bounds of the Confidence Interval are both negative: negative endpoint <p1 - p2< negative endpoint

p1 must be less than p2 by between lower bound and upper bound

The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​ However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of​ indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made.​ Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a​ 95% confidence interval to judge whether the two indenters result in different measurements. 'diamond minus steel ball'

sample 1 = diamond sample 2 = steel ball lower = .3 upper = 2.8 There is sufficient evidence to conclude that the two indenters produce different hardness readings.

Assume that both populations are normally distributed Test whether mu 1 not equals mu 2 at the alpha equals 0.01 level of significance for the given sample data.

u1 = u2 u1 /= u2 p = .291 Do not reject H0​, there is not sufficient evidence to conclude that the two populations have different means. use .99 conf -4.72 and 2.11

A study was done using a treatment group and a placebo group. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.01 significance level for both parts.

u1 = u2 u1 /= u2 t = -1.87 p-val = .067 Fail to reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations with the same mean. -.777 < u1 - u2 < .137

A study was done on proctored and nonproctored tests. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below.

u1 = u2 u1 < u2 -1.82; .038 Fail to reject H0. There is not sufficient evidence to support the claim that students taking proctored tests get a lower mean score than those taking nonproctored tests [use .98 conf] -17.883 < u1 - u2 < 2.503 Yes, because the confidence interval contains zero.

a. Test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. b. Construct a confidence interval suitable for testing the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests.

u1 = u2 u1 < u2 Fail to reject H0. There is not sufficient evidence to support the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests. [use .98] -17.89 < u1 - u2 < 3.47 Yes, because the confidence interval contains zero.

a. Use a 0.05 significance​ level, and test the claim that the mean skull breadth in 4000 B.C is less than the mean skull breadth in A.D 150.

u1 = u2 u1 < u2 Reject the null hypothesis. There is sufficient evidence to support the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150. [use .90 conf] Yes, because the confidence interval contains only negative values.

a. Use a 0.05 significance​ level, and test the claim that the mean skull breadth in 4000 B.C is less than the mean skull breadth in A.D 150..

u1 = u2 u1 < u2 t-stat = -1.22 p-val = .114 Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150. -3.98 < u1 - u2 < .625 Yes, because the confidence interval contains zero.

A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.05 significance level for both parts.

u1 = u2 u1 > u2 Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that men have a higher mean body temperature than women. [use .90 conf] Yes, because the confidence interval contains zero.

Listed in the data table are IQ scores for a random sample of subjects with medium lead levels in their blood. Also listed are statistics from a study done of IQ scores for a random sample of subjects with high lead levels. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.05 significance level for both parts.

u1 = u2 u1 > u2 Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that subjects with medium lead levels have higher IQ scores. [use .9] Yes, because the confidence interval contains zero.

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.

u1 = u2 u1 > u2 p-val = .39 Do not reject H0​, there is not sufficient evidence at the a=0.01 level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms.

Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment​ (with magnets) group and the sham​ (or placebo) group. The results are a measure of reduction in back pain. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.05 significance level for both parts. b. Construct a confidence interval suitable for testing the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment.

u1 = u2 u1 > u2 t = .15 p-val = .441 Fail to reject the null hypothesis. There is not sufficient evidence to support the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. Since the sample mean for those treated with magnets is greater than the sample mean for those given a sham​ treatment, it is valid to argue that magnets might appear to be effective if the sample sizes are larger. [use .90 conf] -.622 < u1 - u2 < .742

Listed below are systolic blood pressure measurements​ (mm Hg) taken from the right and left arms of the same woman. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

u_d = 0 u_d /= 0 -3.08, .037 Since the​ P-value is less than the significance​ level, reject the null hypothesis. There is sufficient evidence to support the claim of a difference in measurements between the two arms

Refer to the data set in the accompanying table. Assume that the paired sample data is a simple random sample and the differences have a distribution that is approximately normal. Use a significance level of 0.01 to test for a difference between the weights of discarded paper​ (in pounds) and weights of discarded plastic​ (in pounds).

u_d = 0 u_d /= 0 3.29, .003 Since the​ P-value is less than the significance​ level, reject the null hypothesis. There is sufficient evidence to support the claim that there is a difference between the weights of discarded paper and discarded plastic.

The accompanying data represent the muzzle velocity​ (in feet per​ second) of rounds fired from a​ 155-mm gun. For each​ round, two measurements of the velocity were recorded using two different measuring​ devices, with the accompanying data obtained.

u_d = 0 u_d /= 0 p-val = 1 Do not reject the null hypothesis. There is not sufficient evidence that there is a difference in the measurements of velocity between device A and device B. -.382, .382 We are​ 99% confident there is no significant difference in the mean muzzle volume using the two measuring devices because the interval includes zero

Listed below are ages of actresses and actors at the time that they won an award for the categories of Best Actress and Best Actor. Use the sample data to test for a difference between the ages of actresses and actors when they win the award. Use a 0.10 significance level. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. (use paired)

u_d = 0 u_d /= 0 t = -2.20 p-val =.092 Since the​ P-value is less than the significance​ level, reject the null hypothesis. There is sufficient evidence to support the claim that there is a difference between the ages of actresses and actors when they win the award.

Test if mu Subscript d less than 0 at the alpha equals 0.05 level of significance.

u_d = 0 u_d < 0

To test the belief that sons are taller than their​ fathers, a student randomly selects 6 fathers who have adult male children. She records the height of both the father and son in inches and obtains the accompanying data. Are sons taller than their​ fathers? Use the alpha equals 0.1 level of significance. Note that a normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

u_d = 0 u_d > 0 [since Yi - Xi, put son height as sample 1 and father height as sample 2] p-val = .330 Do not reject H0. There is not sufficient evidence to conclude that sons are taller than their fathers.

When subjects were treated with a​ drug, their systolic blood pressure readings​ (in mm​ Hg) were measured before and after the drug was taken. Results are given in the table below. Assume that the paired sample data is a simple random sample and that the differences have a distribution that is approximately normal. Using a 0.01 significance​ level, is there sufficient evidence to support the claim that the drug is effective in lowering systolic blood​ pressure?

u_d = 0 u_d > 0 t = 1.22 p-val = .124 Since the​ P-value is greater than the significance​ level, fail to reject H0. There is insufficient evidence to support the claim that the drug is effective in lowering systolic blood pressure.

Use the weights of cans of generic soda as sample​ one, and use the weights of cans of the diet version of that soda as sample two. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Construct a 90​% confidence interval estimate of the difference between the mean weight of the cans of generic soda and the mean weight of cans of the diet version of that soda. Does there appear to be a difference between the mean​ weights?

use .90 conf The mean weight for the generic soda appears to be greater than the mean weight for the diet variety because the confidence interval contains only positive values.

A study was done on body temperatures of men and women. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.01 significance level for both parts.

use .98 conf Yes, because the confidence interval contains zero.

Assume that the paired data came from a population that is normally distributed. Using a 0.05 significance level and dequals xminus ​y, find d overbar ​,s Subscript d ​,the t test​ statistic, and the critical values to test the claim that mu Subscript d equals0. (paired sample)

x - y = d sum d = 1 n = 8 d-bar = sum d / n = 0.125 s^2_d = sum(d-d-bar)^2 / n-1 sum(d-dbar)^2 = 72.785 s^2_d = 10.410714 s_d = 3.227 t = (d-bar - u_d) / s_d / sqrt(n) = (0.125-0) / 3.227 / sqrt(8) = 0.110 t_a/2 = +-2.365