Question

A random sample of 10 shipments of stick-on labels showed the following order sizes. 22,485 56,758 59,762 17,671 16,301 12,262 48,307 51,196 47,326 31,943 Click here for the Excel Data File (a) Construct a 95 percent confidence interval for the true mean order size. (Round your standard deviation answer to 1 decimal place and t-value to 3 decimal places. Round your answers to the nearest whole number.) The 95 percent confidence interval to (b) The confidence interval can be made narrower by: increasing the sample size or decreasing the confidence level. increasing the sample size or increasing the confidence level. decreasing the sample size or decreasing the confidence level. decreasing the sample size or increasing the confidence level.

Answer 1

Answer:

a) $36401.1-2.262\frac{18230.58}{\sqrt{10}}=23360.63 \approx 23361$    

$36401.1+2.262\frac{18230.58}{\sqrt{10}}=49441.56 \approx 49442$  

b) increasing the sample size or decreasing the confidence level

Since if we increase the sample size the margin of error would be lower and if we decrease the confidence level the margin of error would be reduced since the critical value t would be lower

Step-by-step explanation:

We have the following data given

22,485 56,758 59,762 17,671 16,301 12,262 48,307 51,196 47,326 31,943

We can calculate the sample mean and deviation with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=36401.1$ represent the sample mean

$\mu$ population mean

s= 18230.58 represent the sample standard deviation

n=10 represent the sample size  

Part a

The confidence interval for the mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=10-1=9$

The  Confidence level is 0.95 or 95%, the significance is $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case would be $t_{\alpha/2}=2.262$

Replacing the info given we got:

$36401.1-2.262\frac{18230.58}{\sqrt{10}}=23360.63 \approx 23361$    

$36401.1+2.262\frac{18230.58}{\sqrt{10}}=49441.56 \approx 49442$    

Part b

The confidence interval can be made narrower:

increasing the sample size or decreasing the confidence level

Since if we increase the sample size the margin of error would be lower and if we decrease the confidence level the margin of error would be reduced since the critical value t would be lower