Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 68 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 4.2. Provide a 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. (Round your answers to two decimal places.)

Answer 1

Answer: (16.65, 18.35)

Explanation:

By considering the given information , we have

`\overline{x}= 17.5` mm

s= 4.2 mm

n= 68

Significance level
`=\alpha=1-0.90=0.10`

Since population standard deviation is not given ,

Formula :

Confidence interval for population mean is given :-

`\overline{x}\ \pm\ t_(\alpha/2)(s)/(√(n))`

, where n = sample size.

`t_(\alpha/2)` = Two-tailed t-value for significance level of
`(\alpha)` and degree of freedom df= n-1.

`s` = sample standard deviation.

Two-tailed t-value for significance level of
`(0.10)` and degree of freedom df= 67:

`t_(\alpha/2\ ,df)=t_(0.05,\ 67)=1.6679`

95% Confidence interval for population mean:

`17.5\ \pm\ (1.6679)(4.2)/(√(68))`

`=17.5\ \pm\ (1.6679)(4.2)/(8.2462)`

`=17.5\ \pm\ (1.6679)(0.5093255)`

`=17.5\ \pm\ 0.8495`

`=(17.5- 0.8495,\ 17.5+ 0.8495 )=(16.6505,\ 18.3495)\approx(16.65,\ 18.35)`

Hence, the 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel. = (16.65, 18.35)