Respuesta :
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%
Step-by-step explanation:
The null hypothesis is:
[tex]H_{0} = 0.9[/tex]
Because tate Farm claims that policyholders have a customer satisfaction rate of greater than 90%.
The alternate hypotesis is:
[tex]H_{1} < 0.9[/tex]
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:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] 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 [tex]\mu = 0.9, \sigma = \sqrt{0.9*0.1}[/tex]
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 [tex]n = 60[/tex]
Fifteen percent of these policyholders said they were not satisfied with the quality of service.
So 100 - 15 = 85% were satisfied, which means that [tex]p = 0.85[/tex]
Test statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.85 - 0.9}{\frac{\sqrt{0.9*0.1}}{\sqrt{60}}}[/tex]
[tex]z = -1.29[/tex]
[tex]z = -1.29[/tex] 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%
