Question

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a. Calculate percent 90 confidence interval estimate of the mean, given a sample size of 40 orders. b. Calculate 9 0 percent confidence interval for the mean, given the sample size of 75 orders. c. Explain the difference. d. Calculate the minimum sample size needed to identify a %90 confidence interval for the mean, assuming a $5.00 margin of error.

Answer 1

a)
`78.25- 1.64 (22.50)/(√(40))= 72.416`

`78.25+ 1.64 (22.50)/(√(40))= 84.084`

b)
`78.25- 1.64 (22.50)/(√(75))= 73.989`

`78.25+ 1.64 (22.50)/(√(75))= 82.511`

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d)
`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

Solving for n we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

And replacing the info we have:

`n=((1.640(22.50))/(5))^2 =54.46 \approx 55`

Explanation:

Part a

For this case we have the following data given

`\bar X = 78.25` represent the sample mean for the customer order totals

`\sigma =22.50` represent the population deviation

`n= 40` represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be
`\alpha=0.1` and
`\alpha/2 = 0.05` and the critical value from the normal standard distirbution would be given by:

`z_(\alpha/2)=1.64`

And the confidence interval is given by:

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

And replacing we got:

`78.25- 1.64 (22.50)/(√(40))= 72.416`

`78.25+ 1.64 (22.50)/(√(40))= 84.084`

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

`78.25- 1.64 (22.50)/(√(75))= 73.989`

`78.25+ 1.64 (22.50)/(√(75))= 82.511`

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

`ME=z_(\alpha/2)(\sigma)/(√(n))` (a)

Solving for n we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

And replacing the info we have:

`n=((1.640(22.50))/(5))^2 =54.46 \approx 55`