{Exercise 9.24} Consider the following hypothesis test: H0: = 18 Ha: 18 A sample
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Question
{Exercise 9.24} Consider the following hypothesis test: H0: = 18 Ha: 18 A sample of 48 provided a sample mean = 17 and a sample standard deviation s = 4.7.
If requires, round your answers to two decimal places.
a. Compute the value of the test statistic (to three decimal places.)
b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value. p-value is between
c. At = .05, what is your conclusion? p-value is H0
d. What is the rejection rule using the critical value?
Reject H0 if t is or t is What is your conclusion? t = ; H0
Explanation / Answer
A) H0 = 18
Ha not equal to 18
test static = (x-mean)/(standard deviation/sqrt(n))
= (17-18)/(4.7/sqrt(48))
= -1.47
b) as the test static = -1.47
the degree of freedom = 48-1 = 47
as this is a two tailed test and alpha = 0.05
therefore from t table the p = 0.1470
c) as the p value is greater then the significance level therefore we will accept the null hypothesis.
d) from the t table the critical region = t<-1.68 and t >1.68
as the test static = -1.47>-1,47 therefore null hypothesis is not rejected.
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