Answer :
TThe 95% confidence interval for the mean spent on books at this university is (160.711, 179.289).
In the given question we have to find the 95% confidence interval for the mean spent on books at this university.
A sample of 50 students was taken from the local university.
So n=50
These students spent an average of $170 on books this semester, with a standard deviation of $25.50.
So x=$170 and σ = $25.50
So the 95% confidence interval for the mean spent on books at this university.
CI=x ± [tex]z_{\alpha /2}[/tex]× σ/√n
Now finding the value of [tex]z_{\alpha /2}[/tex] for 99%=0.99 confidence interval.
[tex]\alpha[/tex] =1−0.99 = 0.01
[tex]\alpha[/tex]/2 = 0.01/2 = 0.005
[tex]z_{\alpha /2}=z_{0.005}[/tex]
[tex]z_{\alpha /2}[/tex] = 2.576
The z critical value at 99% confidence interval is 2.576.
Putting the value
CI=170 ± 2.576× 25.50/√50
CI=170 ± 2.576× 25.50/7.071
CI=170 ± 2.576×3.606
CI=170 ± 9.289
CI=(170 − 9.289, 170 + 9.289)
CI =(160.711, 179.289)
Hence, the 95% confidence interval for the mean spent on books at this university is (160.711, 179.289).
To learn more about confidence interval link is here
brainly.com/question/24131141
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