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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?

User Lavanya
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1 Answer

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Answer with explanation:

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

As per given , we have


H_0: \mu=42000\\\\ H_a:\mu<42000

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=\frac{\overline{x}-\mu}{(s)/(√(n))} , where n= sample size ,
\overline{x} = sample mean , s = sample standard deviation.

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

Then,
t=(40800-42000)/((2250)/(√(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.

User Azizjon Kholmatov
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