Answer to Question #155071 in Statistics and Probability for Ruth

Question

Question 3. A sample of 150 calls to a customer helpline during one week found that callers were kept waiting on average for 16 minutes with s = 8.

(a) Find the margin of error for this result if we use a 95% confidence interval for the length of time all customers during this period are kept waiting.

(b) Interpret for management the margin of error.

(c) If we only need to be 90% confident, does the confidence interval become wider or narrower?

(d) Find the 90% confidence interval.

Answer 1

We have that

$\bar x = 16$

$s=8$

$n=150$

a) degrees of freedom: $df = n-1=149$

For $\alpha/2=0.025$ with 149 df the critical value from the t-distribution table is 1.976

The formula for margin of error:

$E=t_{\frac{\alpha}{2},df}\frac{s}{\sqrt n}=1.976\cdot\frac{8}{\sqrt{150}}=1.291$

Therefore the margin of error for the result if we use 95% confidence interval for the length of time for all customers during this period are kept waiting is 1.291.

b) Interpretation for management of the margin of error.

At 95% confidence level the average length of time for all customers during this period are kept waiting within about 1.29 minutes of the average 16 minutes.

c) At 90% confidence level the interval becomes narrow due to the confidence interval width decreases the reliability of an interval containing less if the range to possibly cover the mean.

d) The formula for the confidence interval:

$$\bar x\pm t_{\frac{\alpha}{2},df}\frac{s}{\sqrt n}$$

For $\alpha/2=0.05$ with 149 df the critical value from the t-distribution table is 1.655

$16\pm 1.655\cdot\frac{8}{\sqrt {150}}=16\pm1.08$

At 90% confidence level the length of time for all customers during this period are kept waiting is between 14.92 and 17.08 minutes.