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Use the following scenario for the next 4 questions: Based on the results of an annual customer satisfaction survey of State Farm policyholders, State Farm claims that policyholders have a customer satisfaction rate of greater than 90%. To check the accuracy of this claim, a random sample of 60 State Farm policyholders was asked to rate whether they were satisfied with the quality of customer service. Fifteen percent of these policyholders said they were not satisfied with the quality of service. Can we infer that the satisfaction rate is less than the claim with a level of significance of 5%

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Answer:

Since the pvalue of 0.0985 > 0.05, we cannot infer that the satisfaction rate is less than the claim with a level of significance of 5%

Explanation:

The null hypothesis is:


H_(0) = 0.9

Because tate Farm claims that policyholders have a customer satisfaction rate of greater than 90%.

The alternate hypotesis is:


H_(1) < 0.9

Because of the question: Can we infer that the satisfaction rate is less than the claim with a level of significance of 5%.

The test statistic is:


z = (X - \mu)/((\sigma)/(√(n)))

In which X is the sample mean,
\mu is the value tested at the null hypothesis,
\sigma is the standard deviation and n is the size of the sample.

State Farm claims that policyholders have a customer satisfaction rate of greater than 90%.

This means that
\mu = 0.9, \sigma = √(0.9*0.1)

To check the accuracy of this claim, a random sample of 60 State Farm policyholders was asked to rate whether they were satisfied with the quality of customer service.

This means that
n = 60

Fifteen percent of these policyholders said they were not satisfied with the quality of service.

So 100 - 15 = 85% were satisfied, which means that
p = 0.85

Test statistic:


z = (X - \mu)/((\sigma)/(√(n)))


z = (0.85 - 0.9)/((√(0.9*0.1))/(√(60)))


z = -1.29


z = -1.29 has a pvalue of 0.0985.

Since the pvalue of 0.0985 > 0.05, we cannot infer that the satisfaction rate is less than the claim with a level of significance of 5%

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