Answer:
Following are the solution to these choices:
Step-by-step explanation:
[tex]\mu_{0}=39\\\\n = 25\\\\s = 3.8\\\\\bar{X}=40\\\\\alpha =0.1[/tex]
The test assumption is:
[tex]\text{Null Hypothesis} \to H_0:\mu =\mu_{0}\\\\\text{Alternate Hypothesis} \to H_1:\mu >\mu_{0}\\\\[/tex]
It is a checked right-tail, since the alternative hypothesis is produced to classify the argument when the data difference is greater than 0.
[tex]\text{Critical value} =t_{\alpha,n-1} = t_{0.1, 24} =1.3178\\\\\text{Rejection Region:} t_{0} > t_{\alpha,n-1}\\\\\text{Since} \ t_{0} = 1.3158 < 1.3178 = t_{0.1}\\\\[/tex]
The null hypothesis should be rejected: [tex]H_{o}: \mu = 39.0\ at\ \alpha =0.1.[/tex]
They have little enough proof that perhaps the average number of calls per person per week amounts to even more than 39.
