What’s the reciprocal of 5 and 11/12
2 answers:
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Answer:
1. t - statistics = 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.)
t - statistics = 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.