Chp 9 RA, HW, SA

4. A sample of pulse rates of men and women are used to construct a​ 95% confidence interval for the difference between the two population​ means, and the result is negative 12.2 less than mu 1 minus mu 2 less than minus 1.6​, where pulse rates of men correspond to population 1 and pulse rates of women correspond to population 2. Express the confidence interval with pulse rates of women being population 1 and pulse rates of men being population 2.

1.6 < mu 1 - mu 2 < 12.2. Reversing the designation of the samples changes the signs of the endpoints of the interval estimate. The confidence interval using the new designation is 1.6 less than mu 1 minus mu 2 less than 12.2.

2. Which of the following is NOT true when investigating two population​ proportions?

A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.

3. Which of the following is NOT true of confidence interval estimates of the difference between two population​ proportions?

A confidence interval is used to test a claim about two population proportions.

5. Rhino viruses typically cause common colds. In a test of the effectiveness of​ echinacea, 41 of the 47 subjects treated with echinacea developed rhinovirus infections. In a placebo​ group, 97 of the 114 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.

H1 = and H2 not = Use stats crunch proportion stats, two sample, w/ summary-- -get stat test and z score -p val is larger than a so fail to reject, not sufficient evidence to support claim -The​ P-value is greater than the significance level of alpha equals 0.01​, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that echinacea treatment has an effect. -use confidence interval given in stats crunch and then get upper and lower limits -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.

9. 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 0.05 significance​ level, is there sufficient evidence to support the claim that the drug is effective in lowering systolic blood​ pressure?

H1= H2 > open vlaues in stat crunch -Since the​ P-value is less than the significance​ level, reject Upper H 0. There is sufficient evidence to support the claim that the drug is effective in lowering systolic blood pressure.

6. 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.

H1= H2 > stat crunch: stat, t test, 2 summary, w/ summary -find t test and p value -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 -stat crunch and find the CL **if not given: CL= 1-a if includes = and not = CL=1- 2(a) if includes > and <

12. 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.

H1= H2 not =

8. Which of the following is NOT a principle of making inferences from dependent​ samples?

Testing the null hypothesis that the mean difference equals 0 is not equivalent to determining whether the confidence interval includes 0

9. Which of the following is NOT a requirement of testing a claim about the mean of the differences from dependent​ samples?

The degrees of freedom are n-2.

4. Which of the following is NOT a reason why the procedures to estimate differences of two proportions or testing a claim about two proportions​ work?

The form of the confidence interval utilizes the same variance as when testing claims using hypothesis tests.

1. In the largest clinical trial ever​ conducted, 401,974 children were randomly assigned to two groups. The treatment group consisted of​ 201,229 children given the Salk vaccine for​ polio, and the other​ 200,745 children were given a placebo. Among those in the treatment​ group, 33 developed​ polio, and among those in the placebo​ group, 115 developed polio. If we want to use the methods for testing a claim about two population proportions to test the claim that the rate of polio is less for children given the Salk​ vaccine, are the requirements for a hypothesis test​ satisfied? Explain.

The requirements are​ satisfied; the samples are simple random samples that are​ independent, and for each of the two​ groups, the number of successes is at least 5 and the number of failures is at least 5. The requirements for a hypothesis test are as follows. ​1) The sample proportions are from two simple random samples that are independent. Samples are independent if the sample values selected from one population are not related to or somehow naturally paired or matched with the sample values selected from the other population. ​2) For each of the two​ samples, there are at least 5 successes and at least 5 failures. The first condition is met because the children were randomly assigned to the two groups. The second condition is met because at least 5 children developed polio in both​ groups, and at least 5 children did not develop polio in both groups.

1. Which of the following is NOT a requirement of testing a claim or constructing a confidence interval estimate for two population​ portions?

The sample is at least​ 5% of the population.

3. Determine whether the samples are independent or dependent. To test the effectiveness of a drug comma cholesterol levels are measured in 160 men and 160 women after the treatment.

The samples are independent because there is not a natural pairing between the two samples ***Two samples are independent 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. Two samples are dependent if the samples can be paired in some inherent relationship.

5. The form of the confidence interval utilizes the same variance as when testing claims using hypothesis tests.

The two samples are dependent.

6. Which of the following is NOT true when dealing with independent​ samples?

The variance of the differences between two independent random variables equals the variance of the first random variable minus the variance of the second random variable.

10. What design principle is stressed for experiments or observational​ studies?

Using dependent samples with paired data is generally better than using two independent samples.

7. Which of the following is NOT an advantage of pooling sample​ variances?

We often know that sigma 1 equals sigma 2.

7. 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.

h1= h2 > use stat crunch for t-test -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 Does the confidence interval support the conclusion found with the hypothesis​ test? -Yes, because the confidence interval contains zero.

2. In a randomized controlled​ trial, insecticide-treated bednets were tested as a way to reduce malaria. Among 309 infants using​ bednets, 14 developed malaria. Among 272 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. Difference equals ​p(1) minus ​p(2) Estimate for​ difference: -0.0539573 95​% upper bound for​ difference: negative 0.01152504 Test for difference =0 ​(vs < 0): Z= -2.53 P-Value =0.006

look at the little outlined box test stats= -2.53 p val == 0.006 The​ P-value is less than the significance level alpha​, so reject the null hypothesis. There is sufficient evidence to support the claim that the incidence of malaria is lower for infants using bednets. The bednets appear to be effective.

10. 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 minus after" differences. Does hypnotism appear to be effective in reducing​ pain?

open data in stat crunch stat- t-test- paired (fill in info and CI) yes it is effective, because CI doesn't include zero and is entirely greater than zero

8. 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 0.10 significance level to test for a difference between the measurements from the two arms. What can be​ concluded?

open data in stat crunch, stat crunch t stat, paired-- say what sample one and sample 2 represent (can ignore where and group by) p-val is greater than so fail to reject, is not sufficient evidence to support claim

11. 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). In this​ example, mu Subscript d is the mean value of the differences d for the population of all pairs of​ data, where each individual difference d is defined as the weight of discarded paper minus the weight of discarded plastic for a household. What are the null and alternative hypotheses for the hypothesis​ test?

paired t test H1= H2 not = test for difference of two weights-- doesn't say < or > -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.