Here we have:
"\\bar x =410"
"s=20"
"n=25"
significance level "\\alpha=0.05"
"H_0:\\mu\\le400"
"H_a:\\mu>400"
The test is right-tailed.
a) Since the population standard deviation is unknown and the sample size is smaller than 30 we use t statistic.
b) "df=n-1=24"
The critical value for "\\alpha = 0.05" and 24 degrees of freedom is 1.711.
The critical region is "t>1.711"
c) Test staticstic:
Since 2.5 > 1.711 thus t falls in the rejection region we reject the null hypothesis.
d) Then looking up in the Student's t-distribution table with 24 degrees of freedom we find the corresponding probability or p-value.
p-value is 0.0098
Since 0.0098 < 0.05 thus p-value is less than the significance level we reject the null hypothesis.
e) At the 5% significance level the data do provide sufficient evidence to support the claim. The credit manager is 95% confident to conclude that the population mean is not at most GHc400.
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