21.36 Blood pressure. A randomized comparative experiment studied the effect of diet onblood pressure. Researchers divided 54 healthy white males at random into two groups. Onegroup received a calcium supplement, and the other group received a placebo. At thebeginning of the study, the researchers measured many variables on the subjects. The averageseated systolic blood pressure of the 27 members of the placebo group was reported to bex =114.9 with a standard deviation of s=9.3.a. Give a 95% confidence interval for the mean blood pressure of the population fromwhich the subjects were recruited.b. The recipe you used in part (a) requires an important assumption about the 27 men whoprovided the data. What is this assumption?

Answered: Image /qna-images/answer/12471af7-275d-455a-bea9-a5d962ecf7fc.jpg

Answered: 21.36 Blood pressure. A randomized…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =
Scientists are working on a synthetic vaccine (antigen) for the parasitic roundworm. The success of the vaccine hinges on the characteristics of DNA in peptide (protein) produced by the antigen. In the journal, Gene Therapy and Molecular Biology (June, 2009), scientists tested alleles of antigen-produced protein for level of peptide. For a sample of 24 alleles, the mean peptide score was 1.83 and the standard deviation was .15. Use this information to construct a 95% confidence interval for the true mean peptide score in alleles of the antigen-produced protein.
A coach believes that her volleyball team is more athletic than the average volleyball team in America. (Think of that as her research hypothesis.) We know that the average volleyball team in America has a score of 100 on the Standard Test for Athleticism (the STA; not to be confused with the SAT). The coach tests her 25 players and gets an average STA score for them of 109, with a standard deviation of 15. The coach sets her alpha level at .05. Given the results of her study, what decision should she make? Reject the null hypothesis, since she knows that her team plays so well. Reject the null hypothesis, since the mean is above 100 Reject the null hypothesis, since the inferential Z-score is greater than the critical value when alpha is at .05 Do not reject the null hypothesis; the results are likely due to chance factors.
Acne is a common skin disease that affects most adolescents and can continue into adulthood. A study compared the effectiveness of three acne treatments and a placebo, all in gel form, applied twice daily for 12 weeks. The study's 517 teenage volunteers were randomly assigned to one of the four treatments. Success was assessed as clear or almost clear skin at the end of the 12 week period. The results of the study can be seen in the table below. Using the appropriate statistical test, determine if there is significant evidence that the four treatments perform differently. If so, how do they compare.
A researcher wonders whether the recession has changed variability in family size in his city. Before the recession, the mean was 3.7 and the standard deviation was 1.76. The researcher randomly selects 41 families from a population that is normally distributed. His sample has a mean of 3.1 and a standard deviation of 1.45. Test the claim that the standard deviation in size has changed at the αα=.05 significance level.
Sony would like to test the hypothesis that the average age of a PlayStation user is different from the average age of an Xbox user. A random sample of 32 PlayStation users had an average age of 33.7 years while a random sample of 36 Xbox users had an average age of 35.6 years. Assume that the population standard deviation for the age of PlayStation and Xbox users is 3.7 and 3.5 years, respectively. Sony would like to set alpha is 0.10. The critical value for this hypothesis test would be
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Ackerman and Goldsmith (2011) found that students who studied text from printed hardcopy had better test scores than students who studied text presented on the screen. In a related study, a professor noticed that several students in a large class had purchased the e-book version of the course textbook. For the final exam, the overall average for the entire class was μ= 81.7, but the sample of n = 9 students who used e-books had a mean of M = 77.2 with ((SS =392) Is the sample sufficient to conclude that scores for students using e-books were sufficiently different from scores for the regular class? Use a two-tail test with α= .01.(not .05). Please answer the question using all of the steps presented on your practice problem assignment. (null in word, alternative in words, null in symbols, alternative in symbols, critical region t, df, all steps in the analysis computing your computed t, make a decision, and give a conclusion.
A recent study indicates that simply giving college students a pedometer can result inincreased walking (Jackson & Howton, 2008). Students were given pedometers for a 12-week period, and asked to record the average number of steps per day during weeks 1, 6,and 12. The following data are similar to the results obtained in the study. Use a repeatedmeasures ANOVA with α=.05 to determine whether the mean number of steps changessignificantly from one week to another. (do all 4 steps of conducting a hypothesistest!!): If there is a significant different, calculate the percent of variance that is explainedby treatment differences (??2) and conduct both Tukey and Scheffe post hoc tests. Person Week 1 Week 6 Week 12 Person Totals A 6 8 10 P=24 N=18 B 4 5 6 P=15 G=72 C 5 5 5 P=15 D 1 2 3 P=6 E 0 1 2 P=3 F 2 3 4 P=9 M=3 M=4 M=5 T=18 T=24 T=30…
A researcher was interested in comparing the resting pulse rate of people who exercise regularly and people who do not exercise regularly. Independent random samples of 16 people aged 30-40 who do not exercise regularly (sample 1) and 12 people aged 30-40 who do exercise regularly (sample 2) were selected and the resting pulse rate of each person was measured. The summary statistics are as follows: Pulse Rate data Group 1 (no exercise) Group 2 (exercise) average 72.7 69.7 standard deviation 10.9 8.2 sample size 16 12 Test the claim that the mean resting pulse rate of people who do not exercise regularly is greater than the mean resting pulse rate of people who exercise regularly, use 0.01 as the significance level. Round you answer to 3 decimal places. Group of answer choices p-value=0.207, evidence not support claim p-value=0.267, evidence support claim p-value=0.414, evidence not support claim p-value=0.793, evidence not support claim p-value=0.207, evidence…
For a randomized block ANOVA, suppose that the treatment Factor T has four levels and the blocking Factor B has eight levels. Given that SSTR = 900, SSBL = 700, and SSE = 210, construct the randomized block ANOVA table for the data.

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