Question

Here are the weights, in pounds, of a sample of 13 adult female golden retriever dogs: 59.0, 54.1, 53.7, 51.6, 57.5, 58.7, 58.0, 53.8, 48.9, 53.9, 51.6, 55.9, and 57.4. what are the degrees of freedom, and what's the critical value of t needed to construct a 95% confidence interval for the population mean weight of adult female golden retrievers? answer:

Answer 1

Answer:

95% confidence interval for the population mean weight of adult female golden retrievers is [53.01 pounds , 56.78 pounds].

Step-by-step explanation:

We are given that the weights, in pounds, of a sample of 13 adult female golden retriever dogs :

59.0, 54.1, 53.7, 51.6, 57.5, 58.7, 58.0, 53.8, 48.9, 53.9, 51.6, 55.9, and 57.4.

Firstly, the pivotal quantity for 95% 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 weight =  $\frac{\sum X }{n}$  = 54.9 pounds

            s = sample standard deviation =  $\frac{\sum (X-\bar X)^{2} }{n-1}$  = 3.12 pounds

           n = sample of female = 13

           $\mu$ = population mean weight

Here for constructing 95% confidence interval we have used 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(-2.179 < $t_1_2$ < 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 $\mu$ =  $\bar X \pm 2.179 \times {\frac{s}{\sqrt{n} } }$

                                             = [ $54.9 - 2.179 \times {\frac{3.12}{\sqrt{13} } }$ , $54.9 +2.179 \times {\frac{3.12}{\sqrt{13} } }$ ]

                                             = [53.01 pounds , 56.78 pounds]

Therefore, 95% confidence interval for the population mean weight of adult female golden retrievers is [53.01 pounds , 56.78 pounds].