Question

Listed below are student evaluation ratings of​ courses, where a rating of 5 is for​ "excellent." The ratings were obtained at one university in a state. Construct a confidence interval using a 99​% confidence level. What does the confidence interval tell about the population of all college students in the​ state?

3.6​, 3.1​, 4.0​, 4.9​, 3.0​, 4.3​, 3.6​, 4.6​, 4.6​, 4.0​, 4.4​, 3.6​, 3.3​, 4.2​, 3.7


What is the confidence interval for the population mean mu​?

_ < u < _

nothing ​(Round to two decimal places as​ needed.)

Answer 1

`3.93-2.977(0.574)/(√(15))=3.49`

`3.93+2.977(0.574)/(√(15))=4.37`

3.49 < u < 4.37

Explanation:

Data provided

3.6​, 3.1​, 4.0​, 4.9​, 3.0​, 4.3​, 3.6​, 4.6​, 4.6​, 4.0​, 4.4​, 3.6​, 3.3​, 4.2​, 3.7

The sample mean and deviation can be calculated with the following formulas

`\bar X= \sum_(i=1)^n (x_i)/(n)`

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}`

`\bar X=3.93` represent the sample mean

`\mu` population mean

s=0.574 represent the sample standard deviation

n=15 represent the sample size

Confidence interval

The confidence interval for the true mean is given by:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom, given by:

`df=n-1=15-1=114`

The Confidence is 0.99 or 99%, the significance is
`\alpha=0.01` and
`\alpha/2 =0.005`, and the critical value would be
`t_(\alpha/2)=2.977`

Replacing we got:

`3.93-2.977(0.574)/(√(15))=3.49`

`3.93+2.977(0.574)/(√(15))=4.37`

3.49 < u < 4.37