Answer: [tex](82198.72,\ 103801.28)[/tex]
Step-by-step explanation:
Given : Sample size : n=12 , which is small sample (n<30), so we use t-test.
Sample mean : [tex]\overline{x}=93,000[/tex]
Standard deviation : [tex]s=17000[/tex]
Significance level : [tex]\alpha:1-0.95=0.05[/tex]
Then , Critical value : [tex]t_{n-1,\alpha/2}=2.200985[/tex]
The formula to find the confidence interval is given by : -
[tex]\overline{x}\pm t_{n-1,\alpha/2}\dfrac{s}{\sqrt{n}}\\\\\Rightarrow\ 93000\pm (2.200985)\dfrac{17000}{\sqrt{12}}\\\\\approx9,000\pm0.4011=(93000-10801.28,\ 93000+10801.28)=(82198.72,\ 103801.28)[/tex]
Hence, the 95% confidence interval for the average starting salary among all Harvard graduates = [tex](82198.72,\ 103801.28)[/tex]
