Question

Use the given degree of confidence and sample data to construct a confidence interval for the population mean mu. Assume that the population has a normal distribution. Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was​ 6.6, construct a​ 99% confidence interval for the mean score of all students.

Answer 1

Answer:

$95-2.76\frac{6.6}{\sqrt{30}}=91.674$    

$95+2.76\frac{6.6}{\sqrt{30}}=98.326$    

We can say at 99% confidence that the true mean is between (91.674;98.326)    

Step-by-step explanation:

Data given

$\bar X=95$ represent the sample mean

$\mu$ population mean (variable of interest)

s=6.6 represent the sample standard deviation

n=30 represent the sample size  

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

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

We need to find the critical value $t_{\alpha/2}$ and we need to find first the degrees of freedom, given by:

$df=n-1=30-1=29$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,29)".And we see that $t_{\alpha/2}=2.76$

And replacing we got:

$95-2.76\frac{6.6}{\sqrt{30}}=91.674$    

$95+2.76\frac{6.6}{\sqrt{30}}=98.326$    

We can say at 99% confidence that the true mean is between (91.674;98.326)