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 17.5 customer contacts per week. The sample standard deviation was 5.7. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel. 90% Confidence interval, to 2 decimals:

CI for 90% = ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

And,

CI for 95% = ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

Step-by-step explanation:

= ( 2.64, 3.14)

Therefore at 95% confidence interval (a,b) = ( 2.64, 3.14)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 17.5

Standard deviation r = 5.7

Number of samples n = 65

Confidence interval = 90% and 95%

z(at 90% confidence) = 1.645

z(at 95% confidence) = 1.96

Substituting the values we have; for 90%

17.5+/-1.645(5.7/√65)

17.5+/-1.645(0.707)

17.5 +/- 1.16

= ( 16.34, 18.66)

Therefore at 90% confidence interval (a,b) = ( 16.34, 18.66)

Substituting the values we have; for 95%

17.5+/-1.96(5.7/√65)

17.5+/-1.96(0.707)

17.5 +/- 1.39

= ( 16.11, 18.89)

Therefore at 95% confidence interval (a,b) = ( 16.11, 18.89)

the answer is c

step-by-step explanation:

your answer is the frist one

hop i