The Department of Education would like to test - id7158302, montgomerykarloxc24x

Ответ:

mrojas1011

30.01.2022

A 95% confidence interval for the population average debt load of graduating students with a bachelor's degree is [$17,022.05, $20,577.94].

Step-by-step explanation:

We are given that a random sample of 28 students had an average debt load of $18,800. It is believed that the population standard deviation for student debt load is $4800. The α is set to 0.05.

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 average debt load = $18,800

$\sigma$ = population standard deviation = $4,800

n = sample of students = 28

$\mu$ = population average debt load

Here for constructing a 95% confidence interval we have used a One-sample z-test statistics because we know about population standard deviation.

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

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $\$18,800-1.96 \times {\frac{\$4,800}{\sqrt{28} } }$ , $\$18,800+1.96 \times {\frac{\$4,800}{\sqrt{28} } }$ ]

= [$17,022.05, $20,577.94]

Therefore, a 95% confidence interval for the population average debt load of graduating students with a bachelor's degree is [$17,022.05, $20,577.94].

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