Question

A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours. Estimate the mean credit hours taken by a student each quarter using a 95% confidence interval. Round to the nearest thousandth.

Answer 1

To estimate the mean credit hours taken by a student each quarter using a 95% confidence interval, use the formula: Confidence interval = mean ± (critical value * standard deviation / sqrt(sample size)).

Step-by-step explanation:

To estimate the mean credit hours taken by a student each quarter using a 95% confidence interval, you can use the formula:

Confidence interval = mean ± (critical value * standard deviation / sqrt(sample size))

In this case, the critical value can be obtained from the t-distribution table for a 95% confidence level with 249 degrees of freedom.

Plugging in the values, the confidence interval is: 15.2 ± (1.96 * 2.3 / sqrt(250)).

Rounding to the nearest thousandth, the estimated mean credit hours taken by a student each quarter is 15.2 ± 0.289.

Answer 2

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

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 credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

`\mu` = population mean credit hours per quarter

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

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

P(-1.96 <
`t_2_4_9` < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 <
`(\bar X-\mu)/((s)/(√(n) ) )` < 1.96) = 0.95

P(
`-1.96 * {(s)/(√(n) ) }` <
`{\bar X-\mu}` <
`1.96 * {(s)/(√(n) ) }` ) = 0.95

P(
`\bar X-1.96 * {(s)/(√(n) ) }` <
`\mu` <
`\bar X+1.96 * {(s)/(√(n) ) }` ) = 0.95

95% confidence interval for
`\mu` = [
`\bar X-1.96 * {(s)/(√(n) ) }` ,
`\bar X+1.96 * {(s)/(√(n) ) }` ]

= [
`15.2-1.96 * {(2.3)/(√(250) ) }` ,
`15.2+1.96 * {(2.3)/(√(250) ) }` ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.