Question
answer question 19
A student of the author measured the sitting heights of 36 male classmate friends, and she obtained a mean of 92.8 cm. The population of males has sitting heights with a mean of 91.4 cm and a standard deviation of 3, 6 cm (based on anthropometric survey data from Gordon, Churchill, et al.) Use a 0.05 significance level to test the claim that males at her college have a mean sitting height different from 91.4 cm. Is there anything about the sample data suggesting that the methods of this section should not be used? The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 54 bears has a mean weight of 182.9 lb. Assuming that sigma is known to be 121.8 lb, use a 0 05 significance level to test the claim that the population mean of all such bear weights is greater than 150 lb. A simple random sample of 40 salaries of NCAA has a mean of $415, 953. The standard deviation of all salaries of NCAA football coaches is $463.364. Use a 0.05 significance level to test the claim that the mean salary of a football coach in the NCAA is less than $500,000 A sample random sample of 36 cans of regular Coke has a mean volume of 12.19 oz (based on Data Set 17 in Appendix B). Assume that the standard deviation of all cans of regular coke is 0.11 oz. Use a 0.01 significance level to test the claim that cans of regular coke have volumes with mean of 12 oz, as stated on the label. If there is a difference, is it substantial? Tests of older baseballs showed that when dropped 24 ft onto a concrete surface, they bounced an average of 235.8 cm. In a test of 40 new baseballs, the bounce heights had a mean of 235.4 cm. Assume that the standard deviation of bounce heights is 4.5 cm (based on data from Brookhaven National Laboratory and USA Today). Use a 0.05 significance level to test the claim that the new baseballs have bounce heights with a mean different from 235.8 cm. Are the new baseballs different? The totals of the individual weights of garbage discarded by 62 households in one week have a mean of 27.443 lb (based on Data Set 22 in Appendix B). Assume that the standard deviation of the weights is 12.458 lb. Use a 0.05 significance level to test the claim that the population of households has a mean less than 30 lb, which is the maximum amount that can be handled by the current waste removal system. Is there any cause for concern? Using Raw Data. In Exercise 19 and 20, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. FICO Credit Scores A simple random sample of FICO credit rating scores is obtained, and the scores are listed below. As of this writing, the mean FICO score was reported to be 678. Assuming the standard deviation of all FICO scores is known to be 58.3, use a 0.05 significance level to test the claim that these sample FICO scores come from a population with a mean equal to 678. 714 715 664 789 818 779 698 836 753 834 693 802 California Speeding Listed below are recorded speeds (in mi/h) of randomly selected cars travelling on a section of Highway 405 in Los Angles (based on data from Sigalert). That part of the highway has a posted speed limit of 65 mi/h. Assume that the standard deviation of speeds is 5.7 mi/h and use a 0.01 significance level to test the claim that the sample is from a population with a mean that is greater than 65 mi/h. 68 68 72 73 65 74 73 72 68 65 65 73 66 73 66 71 68 74 66 74 66 71 65 73
Explanation / Answer
Step1:
Null hypothesis:
H0:mean of scores =678
Alternative Hypothesis
H1: mean of scores 678
Step2:
Choose levl of significance
alpah=0.05
Step3:
Test statistic:
find sample mean xbar =760.9167
and sample standard deviation using excel=58.28372
t=760.9167-678/58.28372/sqrt(12)
t=4.93
Step4:decision rule
if p<0.05 reject null hypothesis
if p>0.05 fail to reject null hypothesis
Step5:p value
Degrees of freedom=n-1=12-1=11
alpha=0.05
The P-Value is .00045.
The result is significant at p < .05
Step6:
Statistcial decision and conclusion:
since p<0.05 reject Null hypothesis
Accept alternative Hypothesis
There is no sufficient statisticcal evidence at 5% level of significance to support the claim that Scores came from a population with mean =678
