Question

Interpret the results. Select the correct choice below and fill in the answer box to complete your choice. ​(Type an integer or a decimal. Do not​ round.) A. nothing​% of all random samples of people from the population will have a mean driving distance to work​ (in miles) that is between the​ interval's endpoints. B. It can be said that nothing​% of the population has a driving distance to work​ (in miles) that is between the​ interval's endpoints. C. With nothing​% ​confidence, it can be said that most driving distances to work​ (in miles) in the population are between the​ interval's endpoints. D. With nothing​% ​confidence, it can be said that the population mean driving distance to work​ (in miles) is between the​ interval's endpoints.

Answer 1

Answer:

With 99 %  confidence, it can be said that the population mean driving distance to work ​ (in miles) is between the​ interval's endpoints [19.91 miles, 31.49 miles] .

Step-by-step explanation:

The complete question is: In a random sample of  six  ​people, the mean driving distance to work was  25.7 miles and the standard deviation was  6.7  miles. Assuming the population is normally distributed and using the​ t-distribution, a  99​%  confidence interval for the population mean  mu  is  left parenthesis 14.7 comma 36.7 right parenthesis  ​(and the margin of error is  11.0​).

Through​ research, it has been found that the population standard deviation of driving distances to work is  5.5 .  Using the standard normal distribution with the appropriate calculations for a standard deviation that is​ known, find the margin of error and construct a  99 ​%  confidence interval for the population mean  mu .

Interpret the results. Select the correct choice below and fill in the answer box to complete your choice.  ​(Type an integer or a decimal. Do not​ round.)  

A.  nothing ​%  of all random samples of  six  people from the population will have a mean driving distance to work​ (in miles) that is between the​ interval's endpoints.

B.  With  nothing ​%  ​confidence, it can be said that most driving distances to work​ (in miles) in the population are between the​ interval's endpoints.

C.  It can be said that  nothing ​%  of the population has a driving distance to work​ (in miles) that is between the​ interval's endpoints.  

D.  With  nothing %  confidence, it can be said that the population mean driving distance to work ​ (in miles) is between the​ interval's endpoints.

We are given that in a random sample of  six  ​people, the mean driving distance to work was  25.7 miles and the standard deviation was  6.7  miles.

Through​ research, it has been found that the population standard deviation of driving distances to work is  5.5 .

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

                             P.Q.  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~  N(0,1)  

where, $\bar X$ = sample mean driving distance to work = 25.7 miles

             $\sigma$ = population standard deviation = 5.5 miles

            n = sample of people = 6

             $\mu$ = population mean driving distance to work

Here for constructing a 99% confidence interval we have used a One-sample z-test statistics because we know about the population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                      of significance are -2.58 & 2.58}  

P(-2.58 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ]

                                            = [ $25.7-2.58 \times {\frac{5.5}{\sqrt{6} } }$ , $25.7+2.58 \times {\frac{5.5}{\sqrt{6} } }$ ]

                                            = [19.91, 31.49]

Therefore, a 99% confidence for the population mean is [19.91, 31.49] .

The margin of error here is = $2.58 \times {\frac{\sigma}{\sqrt{n} } }$

                                             = $2.58 \times {\frac{5.5}{\sqrt{6} } }$  = 5.793

With 99 %  confidence, it can be said that the population mean driving distance to work ​ (in miles) is between the​ interval's endpoints [19.91, 31.49] .