<h2>
Answer with explanation:</h2>
Let
be the average starting salary ( in dollars).
As per given , we have
![H_0: \mu=42000\\\\ H_a:\mu](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%3D42000%5C%5C%5C%5C%20H_a%3A%5Cmu%3C42000)
Since
is left-tailed , so our test is a left-tailed test.
WE assume that the starting salary follows normal distribution .
Since population standard deviation is unknown and sample size is small so we use t-test.
Test statistic :
, where n= sample size ,
= sample mean , s = sample standard deviation.
Here , n= 15 ,
, s= 225
Then, ![t=\dfrac{40800-42000}{\dfrac{2250}{\sqrt{15}}}\approx-2.07](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B40800-42000%7D%7B%5Cdfrac%7B2250%7D%7B%5Csqrt%7B15%7D%7D%7D%5Capprox-2.07)
Degree of freedom = n-1=14
The critical t-value for significance level α = 0.01 and degree of freedom 14 is 2.62.
Decision : Since the absolute calculated t-value (2.07) is less than the critical t-value., so we cannot reject the null hypothesis.
Conclusion : We do not have sufficient evidence at 1 % level of significance to support the claim that the average starting salary of the graduates is significantly less that $42,000.