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

Answer:

$3.93-2.977\frac{0.574}{\sqrt{15}}=3.49$    

$3.93+2.977\frac{0.574}{\sqrt{15}}=4.37$

3.49 < u < 4.37

Step-by-step 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

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

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

$\bar X=3.93$ 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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:

$3.93-2.977\frac{0.574}{\sqrt{15}}=3.49$    

$3.93+2.977\frac{0.574}{\sqrt{15}}=4.37$

3.49 < u < 4.37