Answer:
Step-by-step explanation:
Hello!
The study variable is:
X: customer contacts made during a week
To estimate the population mean a sample of n= 65 weekly reports with sample X[bar]= 19.5 contacts and standard deviation S= 5.2
To estimate the mean of a population, you need it to be at least normal. There is no information about the variable distribution but since the sample size is large enough, n≥30, you can apply the Central Limit Theorem and approximate the sampling distribution to normal: X[bar]≈N(μ;σ²/n)
With this you can use the approximation to standard normal distribution to estimate the population mean per confidence interval:
X[bar] ± [tex]Z_{1-\alpha /2}[/tex] * [tex]\frac{S}{\sqrt{n} }[/tex]
90% CI
[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]
19.5 ± 1.648 * [tex]\frac{5.2}{\sqrt{65} }[/tex]
[18,44; 20,56]
With a confidence level of 90% you'd expect that the interval [18,44; 20,56] includes the population mean of curstomer contacts made during a week.
95% CI
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
19.5 ± 1.965 + [tex]\frac{5.2}{\sqrt{65} }[/tex]
[18.23; 20,77]
With a confidence level of 95% you'd expect that the interval [18.23; 20,77] includes the population mean of curstomer contacts made during a week.
I hope it helps!
