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?

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. =  ~

where, = sample mean weight =    = 54.9 pounds

            s = sample standard deviation =    = 3.12 pounds

           n = sample of female = 13

            = 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, is ;

P(-2.179 < < 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 < < 2.179) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for =  

                                             = [ , ]

                                             = [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].

the answer is c)8

step-by-step explanaction:

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y