Answer :
Answer:
e) None of the above
Step-by-step explanation:
We are in posession of the sample's standard deviation, so we use the student t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 43 - 1 = 42
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 42 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975[/tex]. So we have T = 2.0181
The margin of error is:
M = T*s = 2.0181*11.3 = 22.80
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 77.86 - 22.8 = 55.06
The upper end of the interval is the sample mean added to M. So it is 77.86 + 22.8 = 100.66.
So the correct answer is:
e) None of the above
Answer:
[tex]77.86-2.02\frac{11.30}{\sqrt{43}}=74.38[/tex]
[tex]77.86+2.02\frac{11.30}{\sqrt{43}}=81.34[/tex]
And for this cae none of the options satisfy the result so then the best option would be:
e) None of the above
Step-by-step explanation:
Information given by the problem
[tex]\bar X= 77.86[/tex] represent the sample mean for the score
[tex]\mu[/tex] population mean
s=11.30 represent the sample standard deviation
n=43 represent the sample size
Calculating the confidence interval
The confidence interval for the true mean of interest is given by:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
For this case the degrees of freedom are:
[tex]df=n-1=43-1=42[/tex]
The Confidence is 0.95 or 95%, the significance then is [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case would be [tex]t_{\alpha/2}=2.02[/tex]
And replacing in equation (1) we got:
[tex]77.86-2.02\frac{11.30}{\sqrt{43}}=74.38[/tex]
[tex]77.86+2.02\frac{11.30}{\sqrt{43}}=81.34[/tex]
And for this cae none of the options satisfy the result so then the best option would be:
e) None of the above