Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 contacts per week. The sample standard deviation was 5.2. Provide a 90% confidence interval for the mean number of weekly customer contacts. (Round to two decimal places) Answer 18.42 and Answer 20.58 Provide a 95% confidence interval for the mean number of weekly customer contacts. (Round to two decimal places) 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] ± *

90% CI

19.5 ± 1.648 *

[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

19.5 ± 1.965 +

[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!

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: