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

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

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. =
`(\bar X -\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

where,
`\bar X` = sample mean weight =
`(\sum X )/(n)` = 54.9 pounds

s = sample standard deviation =
`(\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 <
`(\bar X -\mu)/((s)/(√(n) ) )` < 2.179) = 0.95

P(
`-2.179 * {(s)/(√(n) ) }` <
`{\bar X -\mu}` <
`2.179 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-2.179 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X +2.179 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` =
`\bar X \pm 2.179 * {(s)/(√(n) ) }`

= [
`54.9 - 2.179 * {(3.12)/(√(13) ) }` ,
`54.9 +2.179 * {(3.12)/(√(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].