Question

The accompanying data represent the total travel tax​ (in dollars) for a​ 3-day business trip in 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
68.87 78.25 70.44 84.67 79.79 86.33 100.24 98.26
(a) Determine a point estimate for the population mean travel tax
A point estimate for the population mean travel tax is $ 83.36. (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 lower bound is $ and the upper bound is $. One can be % confident that all cities have a travel tax between these values.
B. The lower bound is $ and the upper bound is $ The travel tax is between these values for % of all cities.
C. The lower bound is $ and the upper bound is $ There is a % probability that the mean travel tax for all cities is between these values.
D. The lower bound is $ and the upper bound is One can be [95]% confident that the mean travel tax for all cities is between these values.
(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?
A. The researcher could decrease the level of confidence.
B. The researcher could decrease the sample standard deviation.
C. The researcher could increase the level of confidence
D. The researcher could increase the sample mean

Answer 1

Answer:

(a) The point estimate for the population mean travel tax is $ 83.36.

(b) The lower bound is $73.70 and the upper bound is $93.02 One can be [95]% confident that the mean travel tax for all cities is between these values.

(c) The researcher could decrease the level of confidence.

Step-by-step explanation:

We are given that a normal probability plot suggests the data could come from a population that is normally distributed.

X: 68.87, 78.25, 70.44, 84.67, 79.79, 86.33, 100.24, 98.26.

(a) A point estimate for the population mean travel tax is the sample mean of the data. i.e;

      Point estimate, $\bar X$ =  $\frac{\sum X}{n}$

                                     =  $\frac{68.87+ 78.25+ 70.44+ 84.67+ 79.79+ 86.33+ 100.24+ 98.26}{8}$

                                     =  $\frac{666.85}{8}$  = $83.36

So, the point estimate for the population mean travel tax is $ 83.36.

(b) Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                             P.Q.  =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean travel tax = $83.36

             s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$  = $11.55

             n = sample size = 8

             $\mu$ = population mean travel tax

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.365 < N(0,1) < 2.365) = 0.95  {As the critical value of t at 7 degrees of

                                                 freedom are -2.365 & 2.365 with P = 2.5%}  

P(-2.365 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.365) = 0.95

P( $-2.365 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.365 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.365 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.365 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.365 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.365 \times {\frac{s}{\sqrt{n} } }$ ]

                        = [ $\ 83.36-2.365 \times {\frac{\ 11.55}{\sqrt{8} } }$ , $\ 83.36+2.365 \times {\frac{\ 11.55}{\sqrt{8} } }$ ]

                        = [$73.70, $93.02]

Therefore, a 95% confidence interval for the population average debt load of graduating students with a bachelor's degree is [$73.70, $93.02].

The lower bound is $73.70 and the upper bound is $93.02 One can be [95]% confident that the mean travel tax for all cities is between these values.

(c) The researcher could decrease the level of confidence who wants to increase the precision of the interval but does not have access to additional data.