Intro Stats, Books a la Carte Edition (5th Edition)
5th Edition
ISBN: 9780134210285
Author: Richard D. De Veaux, Paul Velleman, David E. Bock
Publisher: PEARSON
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Chapter 16, Problem 16E
To determine
Explain the decision based on the
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p 15 Is sunscreen with an SPF 30 rating more effective at preventing sunburn than sunscreen with an SPF 15 rating? At the alpha = 0.05 level of significance, test the claim that SPF 30 sunscreen is better at preventing sunburn than 15 sunscreen. Let represent the proportion of SPF 15 sunscreen wearers who get a sunburn and represent the proportion of SPF 30 sunscreen wearers who get a sunburn. (Round your results to three decimal places)
1. Which would be correct hypotheses for this test? O O H 0 :p 15 =p 30 , H 1 :p 15 ne p 30 H 0 :p 15 =p 30 , H H 1 :p 15 <p 30 O H 0 :p 15 <p 30 ,H 1 :p 15 >p 30 O H 0 :p 15 =p 30 H 1 :p 15 >p 30
A random sample of 250 beach-goers were given SPF 15 sunscreen and asked to lay in the sun for 1 hour10 of them later reported a sunburn. Another random sample of 250 beach- goers were given SPF 30 sunscreen and asked to lay in the sun for 1 hour. 7 of them later reported a sunburn.
2.Find the test statistic
3. Give the P-value:
answer only…
A question of medical interest is whether jogging leads to a reduction in systolic blood pressure. To learn about this question, eight non-jogging volunteers have agreed to begin a 1-month jogging program. At the end of the month, their blood pressures were determined and compared with earlier values. The data are presented in the Table 1.
a. The appropriate hypotheses are:
H0 :μ1−μ2 =0 versus Ha :μ1−μ2 /= (does not equal) 0
H0 :μ1−μ2 =0 versus Ha :μ1−μ2 <0
H0 :μ1−μ2 =0 versus Ha :μ1−μ2 >0
b. The value of test statistic is
1. 0.577
2. −0.577
3. 0.237
4. -0.237
c. The p−value of test is
1. 0.816
2. 0.408
3. 0.582
4. 0.291
d. At the significance level calculated in part (c), we conclude that jogging
i. Leads to reduction in systolic blood pressureii. Does not lead to reduction in systolic blood pressure
Infants, even newborns, prefer to look at attractive faces compared to less attractive faces (Slater, et al., 1998). In the study, infants from 1 to 6 days old were shown two photographs of women’s faces. Previously, a group of adults had rated one of the faces as significantly more attractive than the other. The babies were positioned in front of a screen on which the photographs were presented. The pair of faces remained on the screen until the baby accumulated a total of 20 seconds of looking at one or the other. The number of seconds looking at the attractive face was recorded for each infant. Suppose that the study used a sample of n = 9 infants and the data produced an average of M = 13 for the attractive face with an estimated standard error sM= 1 (SS = 72). If there were no preference, the 20 seconds should be divided equally between the two photographs. Note that all the available information comes from the sample. Specifically, we do not know the population mean or the…
Chapter 16 Solutions
Intro Stats, Books a la Carte Edition (5th Edition)
Ch. 16.2 - An experiment to test the fairness of a roulette...Ch. 16.2 - Prob. 2JCCh. 16.2 - Prob. 3JCCh. 16.3 - Prob. 4JCCh. 16.3 - Prob. 5JCCh. 16.3 - Prob. 6JCCh. 16.3 - Prob. 7JCCh. 16.4 - Remember the bank thats sending out DVDs to try to...Ch. 16.4 - Prob. 9JCCh. 16.4 - For the bank, which situation has higher power: a...
Ch. 16 - True or false Which of the following are true? If...Ch. 16 - False or true Which of the following are true? If...Ch. 16 - P-values Which of the following are true? If...Ch. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - More critical values For each of the following...Ch. 16 - Prob. 9ECh. 16 - Significant again? A new reading program may...Ch. 16 - SECTION 16.4 11. Errors For each of the following...Ch. 16 - More errors For each of the following situations,...Ch. 16 - CHAPTER EXERCISES 13. P-value A medical researcher...Ch. 16 - Prob. 14ECh. 16 - Alpha A researcher developing scanners to search...Ch. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Is the Euro fair? Soon after the Euro was...Ch. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Alzheimers Testing for Alzheimers disease can be a...Ch. 16 - Prob. 29ECh. 16 - Quality control Production managers on an assembly...Ch. 16 - Cars, again As in Exercise 29, state regulators...Ch. 16 - Prob. 32ECh. 16 - Equal opportunity? A company is sued for job...Ch. 16 - Stop signs Highway safety engineers test new road...Ch. 16 - Prob. 35ECh. 16 - Ads A company is willing to renew its advertising...Ch. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Catheters During an angiogram, heart problems can...Ch. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Faulty or not? You are in charge of shipping...Ch. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48E
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- Repeat Example 5 when microphone A receives the sound 4 seconds before microphone B.arrow_forwardInfants, even newborns, prefer to look at attractive faces compared to less attractive faces (Slater, et al., 1998). In the study, infants from 1 to 6 days old were shown two photographs of women’s faces. Previously, a group of adults had rated one of the faces as significantly more attractive than the other. The babies were positioned in front of a screen on which the photographs were presented. The pair of faces remained on the screen until the baby accumulated a total of 20 seconds of looking at one or the other. The number of seconds looking at the attractive face was recorded for each infant. Suppose that the study used a sample of n = 9 infants and the data produced an average of M = 13 for the attractive face with an estimated standard error sM= 1 (SS = 72). If there were no preference, the 20 seconds should be divided equally between the two photographs. Note that all the available information comes from the sample. Specifically, we do not know the population mean or the…arrow_forwardSetting the alpha value at α = .10 instead of α = .05 increases the chances of: A. making a Type I Error B. making both a Type I and Type II Error C. decreasing power D. making a Type II Errorarrow_forward
- Arsenic-based additives in chicken feed have been banned by the European Union. Ifa restaurant chain finds significant evidence that the mean arsenic level of theirchickens is above 80 ppb (parts per billion), the chain will stop using that supplier ofchicken meat. The hypotheses are: H 0 : µ = 80H 1 : µ > 80 where µ represents the mean arsenic level in all chicken meat from that supplier.Samples from two different suppliers are analyzed, and the resulting p-values aregiven: Sample from Supplier A: p-value is 0.0003Sample from Supplier B: p-value is 0.3500 a) Interpret each p-value in terms of the probability of the results happening byrandom chance. b) Which p-value shows stronger evidence for the alternative hypothesis? c) Which supplier, A or B, should the chain get chickens from in order to avoid toohigh a level of arsenic?arrow_forwardA researcher is testing the effect of marijuana use on reaction time. A sample of n = 9 students is obtained and each student is given a dose of marijuana. Thirty minutes later, each student's reaction time is measured in a driving simulator. The scores for the sample of students averaged M = 211 milliseconds with SS = 1152. Assuming that reaction time for students in the regular population averages μ = 200 milliseconds, are the data sufficient to conclude that the medication has a significant effect on reaction time? Test at the .05 level of significance. Which of the four options below is the best answer? Type the letter “A” or “B” etc. in the box Option A. Accept the null, marijuana has no effect on reaction time Option B. Accept the null, marijuana does affect reaction time Option C. Reject the null, marijuana has no effect on reaction time Option D. Reject the null, marijuana does affect reaction time Compute r 2, the percentage of variance explained by the treatment effect.…arrow_forwardalpha here is 0.5this question is under statistical inferencearrow_forward
- When the necessary conditions are met, you are testing the hypotheses H0: (μ1 - μ2) = 0; H1: (μ1 - μ2) < 0. But your statistical software provides only a one-tail area of 0.023 as part of its output. The p-value for this test will be: A. .023 B. .046 C. .092 D. .082arrow_forwardSetting a smaller alpha level means: A) No answer text provided. B)The acceptance region becomes smaller C) The value of the critical region becomes more positive/negative D) The p value should be large enough to reject the null hypothesisarrow_forwardYou would like to test whether race (Black or White) is associated with one's choice of political party (Democrat, Independent, or Republican). If the Chi-square test statistic you calculate is 234.73, and your alpha = 0.05, what can you conclude? Choices of answers: A: there is no evidence of an association between race and political party B: there is evidence of an association between race and political parties. C: WHite individuals are more likely to identify as Democrats than Black individuals D: Black individuals are more likely to identify as Democrats than White individualarrow_forward
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