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 following parts (a) through (c) below 67.71 78.17 70.26 84.51 79.66 86.41 100.73 98.34 click the icon to view the table of critical t - values (a) determine a point estimate for the population mean travel tax a point estimate for the population mean travel tax is $ (round to two decimal places as needed.) (b) construct and interpret a 95% confidence interval for the mean tax paid for a three - day business trip select the correct choice below and fill in the answer boxes to complete your choice (round to two decimal places as needed.) a. the travel tax is between $ and $ for % of all cities b. one can be % confident that the mean travel tax for all cities is between $ and $ c. there is a % probability that the mean travel tax for all cities is between $ and $ d. one can be % confident that the all cities have a travel tax between $ and $ (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?

Explanation:

Step1: Calculate the point estimate (sample mean)

The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 8\), and \(x_{1}=67.71,x_{2}=78.17,x_{3}=70.26,x_{4}=84.51,x_{5}=79.66,x_{6}=86.41,x_{7}=100.73,x_{8}=98.34\)
\(\sum_{i=1}^{8}x_{i}=67.71 + 78.17+70.26+84.51+79.66+86.41+100.73+98.34=665.79\)
\(\bar{x}=\frac{665.79}{8}=83.22375\approx83.22\)

Step2: Calculate the sample standard deviation \(s\)

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((x_{1}-\bar{x})^{2}=(67.71 - 83.22)^{2}=(-15.51)^{2}=240.5601\)
\((x_{2}-\bar{x})^{2}=(78.17 - 83.22)^{2}=(-5.05)^{2}=25.5025\)
\((x_{3}-\bar{x})^{2}=(70.26 - 83.22)^{2}=(-12.96)^{2}=167.9616\)
\((x_{4}-\bar{x})^{2}=(84.51 - 83.22)^{2}=(1.29)^{2}=1.6641\)
\((x_{5}-\bar{x})^{2}=(79.66 - 83.22)^{2}=(-3.56)^{2}=12.6736\)
\((x_{6}-\bar{x})^{2}=(86.41 - 83.22)^{2}=(3.19)^{2}=10.1761\)
\((x_{7}-\bar{x})^{2}=(100.73 - 83.22)^{2}=(17.51)^{2}=306.6001\)
\((x_{8}-\bar{x})^{2}=(98.34 - 83.22)^{2}=(15.12)^{2}=228.6144\)
\(\sum_{i = 1}^{8}(x_{i}-\bar{x})^{2}=240.5601+25.5025 + 167.9616+1.6641+12.6736+10.1761+306.6001+228.6144=993.7525\)
\(s=\sqrt{\frac{993.7525}{8 - 1}}=\sqrt{\frac{993.7525}{7}}\approx11.91\)

Step3: Find the critical value \(t_{\alpha/2}\)

For a \(95\%\) confidence interval, \(\alpha=1 - 0.95 = 0.05\), and \(\frac{\alpha}{2}=0.025\), \(n=8\), \(df=n - 1=7\)
From the \(t\) - distribution table, \(t_{0.025,7}=2.365\)

Step4: Calculate the margin of error \(E\)

The formula for the margin of error \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\)
\(E = 2.365\times\frac{11.91}{\sqrt{8}}\)
\(E=2.365\times\frac{11.91}{2.828}\approx2.365\times4.21\approx9.96\)

Step5: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)
\(83.22-9.96 <\mu<83.22 + 9.96\)
\(73.26<\mu<93.18\)

Answer:

(a) A point estimate for the population mean travel tax is \(\$83.22\)
(b) B. One can be \(95\%\) confident that the mean travel tax for all cities is between \(\$73.26\) and \(\$93.18\)
(c) Decrease the confidence level.