Question

quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.54. Assume the underlying population is normally distributed. Find the 95% confidence interval for the true population mean for the amount of soda served. (Round your answers to two decimal places.) ,

Answer 1

Answer:

95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].

Step-by-step explanation:

We are given that quality control specialist for a restaurant chain takes a random sample of size 13 to check the amount of soda served in the 16 oz. serving size.

The sample mean is 13.30 with a sample standard deviation of 1.54.

Firstly, the pivotal quantity for 95% confidence interval for the true population mean is given by;

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

where, $\bar X$ = sample mean = 13.30

            s = sample standard deviation = 1.54

            n = sample size = 13

            $\mu$ = true population mean

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

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

P(-2.179 < $t_1_2$ < 2.179) = 0.95  {As the critical value of t at 12 degree of

                                            freedom are -2.179 & 2.179 with P = 2.5%}  

P(-2.179 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.179) = 0.95

P( $-2.179 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.179 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.179 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.179 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.179 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.179 \times {\frac{s}{\sqrt{n} } }$ ]

                                                 = [ $13.30-2.179 \times {\frac{1.54}{\sqrt{13} } }$ , $13.30+2.179 \times {\frac{1.54}{\sqrt{13} } }$ ]

                                                = [12.37 , 14.23]

Therefore, 95% confidence interval for the true population mean for the amount of soda served is [12.37 , 14.23].