Please check my answer!
2 answers:
You might be interested in
Answer:
Given:
mean, u = 6.2
sample size, n = 180
Sample mean, X' = 6.3
s.d = 0.9
Significance level = 0.05
The null and alternative hypothesis will be:
H0 : u = 6.2
H1 : u > 6.2
Degree of freedom = 180 - 1 = 179
Using t table, the t critical value,
t> t(0.05, 179) = 1.6534
The test statistic:
Since the test statistic(t calculated value) 1.4907 < t critical value (1.6534), we fail to reject the null hypothesis H0.