Respuesta :
Answer:
1. t - statistics = Â 1.581
2. We accept null hypothesis that  the mean number of calls is less than or equal to, i.e., ≤, 41 per week.
Step-by-step explanation:
Significance level = 0.025
Sample size = 38
Sample mean = 42
Standard deviation = 3.9
Claimed mean = 41
1. Compute the value of the test statistic. (Round your answer to 3 decimal places.)
[tex]t-statistics = \frac{42 - 41}{3.9/\sqrt{38} }[/tex]
t - statistics = Â 1.581
2. What is your decision regarding null hypothesis? The mean number of calls is than 41 per week.
Degree of freedom = Sample size - 1 = 38 - 1 = 37
At 0.025 significance level,
t-table = 2.021
Since t - statistics (i.e. 1.581) is less than t-table (2.021), we accept null hypothesis that  the mean number of calls is less than or equal to, i.e., ≤, 41 per week.
Answer:
Step-by-step explanation:
For the null hypothesis,
H0 : μ ≤ 41
For the alternative hypothesis,
H1 : μ > 41
Looking at the alternative hypothesis, this a a right tailed test.
Since no population standard deviation is given, the distribution is a student's t.
Since n = 38,
Degrees of freedom, df = n - 1 = 38 - 1 = 37
t = (x - µ)/(s/√n)
Where
x = sample mean = 42
µ = population mean = 41
s = samples standard deviation = 3.9
The test statistic would be
t = (42 - 41)/(3.9/√38) = 1.58
We would determine the p value using the t test calculator. It becomes
p = 0.28
Since alpha, 0.025 < than the p value, 0.28, then we would fail to reject the null hypothesis. Therefore, At a 2.5% level of significance, the sample data did nor show significant evidence that the mean number of calls is more than 41 per week.
