Question

A group of 59 randomly selected students have a mean score of 29.5 with a standard

deviation of 5.2 on a placement test. What is the 90% confidence interval for the mean

score, ", of all students taking the test?

Answer 1

Answer:

$29.5-1.671\frac{5.2}{\sqrt{59}}=28.37$    

$29.5+1.671\frac{5.2}{\sqrt{59}}=30.63$    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence

Step-by-step explanation:

Information given

$\bar X = 29.5$ represent the sample mean

$\mu$ population mean  

s= 5.2 represent the sample standard deviation

n=59 represent the sample size  

Confidence interval

The confidence interval for the true mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are given by:

$df=n-1=59-1=58$

The Confidence interval is 0.90 or 90%, the significance is $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical value is $t_{\alpha/2}=1.671$

Now we have everything in order to replace into formula (1):

$29.5-1.671\frac{5.2}{\sqrt{59}}=28.37$    

$29.5+1.671\frac{5.2}{\sqrt{59}}=30.63$    

And we can conclude that the true mean for the scores are given by this interval (28.37; 30.63) at 90% of confidence