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

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

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)   $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step 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:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

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

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

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:

 $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$