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!

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

jgfjttgbgcfbc

step-by-step explanation:

h56

j

jgh

j

gh

hnh

vnf