Question

A survey claims that a college graduate from Smith College can expect an average starting salary of $ 42,000. Fifteen Smith College graduates had an average starting salary of $ 40,800 with a standard deviation of $ 2250. At the 1 % level of significance, can we conclude that the average starting salary of the graduates is significantly less that $42,000?

Answer 1

Answer with explanation:

Let $\mu$ be the average starting salary ( in dollars).

As per given , we have

$H_0: \mu=42000\\\\ H_a:\mu$

Since $H_a$ 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 : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$ , where n= sample size , $\overline{x}$ = sample mean , s = sample standard deviation.

Here , n= 15 , $\overline{x}= 40,800$  , s= 225

Then, $t=\dfrac{40800-42000}{\dfrac{2250}{\sqrt{15}}}\approx-2.07$

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.