Question

In: Statistics and Probability

A community college claims that the standard deviation in the cost of TI calculators is at...

A community college claims that the standard deviation in the cost of TI calculators is at least $15. A simple random sample of 43 stores that sell TI calculators yielded a sample mean of $84 and a sample standard deviation of $12 for the cost of TI calculators. Does this data refute the claim made by the community college? Perform a statistical test at the 5% significance level. You may assume that the cost of TI calculators is normally distributed.

(a) Identify the parameter of interest and the hypotheses.

(b) Identify the test or distribution you will use. Give a brief explanation of your answer.

(c) Calculate the p-value.

(d) Make a decision regarding the null hypothesis.

(e) Interpret your answer to (d) by writing a meaningful conclusion.

Solutions

Expert Solution

Parameter of interest is the cost of TI calculator

To Test :-

H0 :-  

H1 :-  



n = 43


Test Statistic :-


= 26.88


Test Criteria :-
Reject null hypothesis if

= 26.88 < 58.124 , hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0


Decision based on P value
P value = P (   > 26.88 )
P value = 0.9663
Reject null hypothesis if P value <
Since P value = 0.9663 > 0.05, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

Yes, the given data refute the claims that the standard deviation in the cost of TI calculators is at least $15.



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