Question

The accompanying data represent the total travel tax​ (in dollars) for a​ 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts​ (a) through​ (c) below.

68.76; 79.87; 69.25; 83.63; 79.94; 85.95; 101.77; 98.97

(a) Determine a point estimate for the population mean travel tax.

​(b) Construct and interpret a 95​% confidence interval for the mean tax paid for a​ three-day business trip.

(c) What would you recommend to a researcher who wants to increase the precision of the​ interval, but does not have access to additional​ data?

Answer 1

(a)

Point Estimate of the population mean travel tax =

(68.76 + 79.87 + 69.25 + 83.63 + 79.94 + 85.95 + 101.77 + 98.97) / 8

= 83.5175

(b)

Variance of travel tax = [(68.76 - 83.5175)² + (79.87 - 83.5175)² + (69.25 - 83.5175)² + (83.63 - 83.5175)² + (79.94 - 83.5175)² + (85.95 - 83.5175)² + (101.77 - 83.5175)² + (98.97 - 83.5175)² ] / 7

= 146.473

Standard error of mean = = 4.278916

Degree of freedom = n - 1 = 8 - 1 = 7

Critical value of t at df = 7 and 95​% confidence interval is  2.365

Margin of error = t * Std error = 2.365 * 4.278916 = 10.1196

95​% confidence interval for the mean tax paid for a​ three-day business trip is

(83.5175 - 10.1196, 83.5175 + 10.1196)

(73.40,  93.64)

We're 95% confident that the interval (73.40,  93.64) captured the true mean tax paid for a​ three-day business trip.

(c)

The precision of the​ interval increases by increasing the confidence level or the sample size.

If we not have access to additional​ data, we cannot increase the sample size. Thus, to increase the precision of the​ interval we need to use the confidence level of more than 95%.