Question

Consider the following hypothesis test:

Ho: u = 18
Ha: u ≠ 18

A sample of 48 provided a sample mean of 17 and a sample standard deviation of 4.9. Enter negative values as negative numbers.

a. Compute the value of the test statistic (to three decimal places.)  

b. compute a range for the p-value. (to two decimal places

c. At α=.05, what is your conclusion?
p-value is greater than .05, do not reject Ho
d. What is the rejection rule using the critical value?
reject Ho if T is _____________ or t is ____________________
What is your conclusion?
t=________________

Answer 1

Part a

Test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

t = (17 – 18)/[4.9/sqrt(48)]

t = -1/ 0.707254

t = -1.41392

t = -1.414

Part b

Given:

n = 48

df = n – 1 = 48 – 1 = 47

P-value = 0.1640

P-value = 0.16

(by using t-table)

Part c

P-value is greater than α = 0.05, so we do not reject the null hypothesis H₀

Part d

Reject H₀ if T is less than -2.0117 or T is greater than 2.0117

(Critical values are calculated by using t-table or excel)

Test statistic t = -1.414 is lies within above two critical values, so we do not reject the null hypothesis.

Conclusion: There is insufficient evidence to conclude the researchers claim or an alternative hypothesis.