Test the following claim. Identify the null hypothesis, alternative .pdf

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The document tests a claim about a sample of credit scores using a z-test. It calculates the test statistic z to be 5.11 and the p-value to be 0. The null hypothesis that the population mean is 675 is rejected. Therefore, there is sufficient evidence to conclude that the sample credit scores do not come from a population with a mean of 675.

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Test the following claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-
value, conclusion about the null hypothesis, and final conclusion that addresses the original
claim.
A simple random sample of credit rating scores is obtained, and the scores are listed below. The
mean credit sscore was reported to be 675. Assuming the population is normally distributed ant
the standard deviation of all credit scores is known to be 57.6, use a 0.05 significane level to test
the claim that these sample credit scores come from a population with a mean equal to 675.
714, 752, 665, 789, 818, 780, 697, 837, 754, 834, 692, 801.
- What is the value of the test statistic? z= __?__ (Round to two decimal places as needed.)
The P-value is __?__ (Round to four decimal places as needed.)
- Do we fail to reject or reject the Ho? Is there sufficient evidence to warrant rejection of the
claim that these sample credit scores are from a population with a mean credit score equal to
675?
Solution
the value of the test statistic:
Z=(xbar-mu)/(s/vn)
=(761.0833-675)/(58.39592/sqrt(12))
=5.11
The p-value= 2*P(Z>5.11)=0 (from standard normal table)
Reject HO.
there is sufficient evidence to warrant rejection of the claim that these sample credit scores are
from a population with a mean credit score equal to 675

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