Math  /  Data & Statistics

Question20. A company claims that it has a mean customer service score of 78 (out of 100). You examine 150 customer reviews and find that the mean score of those reviews is 73 with a standard deviation of 2 . Test the accuracy of the company's claim at the 2%2 \% significance level.
Proposed Solution: H0:μ=78H_{0}: \mu=78 Ha:μ<78H_{a}: \mu<78 (since xˉ\bar{x} is less than 78 ) Rejection Region: qt(0.02,149); <2.07\mathbf{<}-2.07 Test Statistic: t=(7378)/(2/\mathrm{t}=(73-78) /(2 / sqrt (150))=30.61(150))=-30.61 In rejection region: (RHo) P-Value Test: pt(-30.61,149) =0%=0 \% p<α(RHo)p<\alpha(R H o) The customer value score is lower than the claimed 78. What is wrong with the proposed solution? A. A z-test should have been used instead of t-test. B. p^\hat{p} should be calculated as 78/100=0.7878 / 100=0.78 and a proportion test used. C. The alternate hypothesis is not supported by the question. D. For p<αp<\alpha, the wrong conclusion was drawn. E. There is nothing wrong with the proposed solution.

Studdy Solution
Verify the conclusion based on the p-value:
- The p-value is calculated to be 0%0\%, which is less than the significance level α=2%\alpha = 2\%. - The conclusion that the customer value score is lower than the claimed 78 is correct.
The correct answer is:
E. There is nothing wrong with the proposed solution.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord