Answer with Step-by-step explanation:
Since we have given
n = 16
a) Describe the sampling distribution for the sample mean.
The sampling distribution of the mean is the population's mean from where the items are sampled.
It would be [tex]\mu[/tex]
b. What is the standard error?
[tex]s=\dfrac{\sigma}{\sqrt{n}}\\\\s=\dfrac{14}{\sqrt{16}}\\\\s=\dfrac{14}{4}\\\\s=3.5[/tex]
c. For 90% confidence, what is the margin of error?
Degrees of freedom = v - 1= 16-1 = 15
Using t-table, 1.753 cuts 5% level of significance in each tail.
so, Margin of error would be
[tex]1.753\times 3.5\\\\=6.1355[/tex]
d. Based on the sample results, create the 90% confidence interval and interpret.
So, 90% confidence interval would be
[tex]110+6.136\\\\=116.136\\\\and\\\\110-6.136\\\\=103.86[/tex]
(103.86,116.136)
