Question

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week.

A sample of 75 weekly reports showed a sample mean of 19.5 customer contacts per week.


The sample standard deviation was 5.2.


Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel:


a. 90% Confidence, to 2 decimals


b. 95% Confidence, to 2 decimals

Answer 1

a) The 90% confidence interval would be given by (18.52;20.49)

b) The 95% confidence interval would be given by (18.32;20.68)

Explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=19.5` represent the sample mean for the sample

`\mu` population mean (variable of interest)

`\sigma=5.2` represent the population standard deviation

n=75 represent the sample size

2) Confidence interval 90%

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.005`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that
`z_(\alpha/2)=1.64`

Now we have everything in order to replace into formula (1):

`19.5-1.64(5.2)/(√(75))=18.515`

`19.5+1.64(5.2)/(√(75))=20.485`

So on this case the 90% confidence interval would be given by (18.52;20.49)

3) Confidence interval 95%

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

Now we have everything in order to replace into formula (1):

`19.5-1.96(5.2)/(√(75))=18.323`

`19.5+1.96(5.2)/(√(75))=20.677`

So on this case the 95% confidence interval would be given by (18.32;20.68)