A sample mean, sample size, and sample standard deviation are provided below. Use the one-mean t-test to perform the required hypothesis test at the 5% significance level.
X=25 s=9 n=24 H0: u=24 Ha: u>24
Solution :
This is the two tailed test,
The null and alternative hypothesis is ,
H0 : = 24
Ha : > 24
Test statistic = t =
= ( - ) / s / n
= (25 - 24 ) / 9 / 24
Test statistic = t = 0.54
degrees of freedom = n - 1 = 24 - 1 = 23
P(t > 0.54 ) = 1-P (t < 0.54 ) = 1 - 0.7028
P-value = 0.2972
= 0.05
P-value >
Fail to reject the null hypothesis.
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