6x -2x3 + 10x+2x3+7x4 simply
2 answers:
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<h2>Answer with explanation:</h2>
Let be the average starting salary ( in dollars).
As per given , we have
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,
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.