Question

1. One‐Sample Univariate Hypothesis Testing of a Mean

Consider a random sample of 5 adults over the age of 25 from a large population, which is normally distributed, where E represents the total years of education completed: ? = [10, 12, 12, 16, 16] Suppose that someone claims that the average person in the population is a college graduate (? = 16).

A. What is the null hypothesis?  What is the alternative hypothesis?

B. Can you reject the null hypothesis at the 10‐percent level of significance?   Can you reject the null hypothesis at the 5‐percent level of significance?   Use the critical value approach.  You can use R for critical values, but you must show all of your calculations and explain.  Use R, however, to check your work.

C. What is the 95‐percent confidence interval for years of education?  Provide a written interpretation explaining your answer.

Answer 1

Answer to the question)

Part a)

Null hypothesis; The average person is a a college graduate with total years of education completed = 16

Alternate hypothesis: The average person is NOT a college graduate with total years of education NOT equal to 16

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Part b)

We got the sample E = [10, 12 , 12 , 16 , 16]

Sample mean ( xbar) = 13.2

Sample standard deviation ( s ) = 2.68

[use the manual formulae, or excel or online calculator to find these values]

Sample size n = 5

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Formula of test statistic

T = (x bar – M) / (s / sqrt(n))
T = (13.2 -16) / (2.68 /sqrt(5))

T = -2.34.

For critical value at significance level 0.10, two tailed test with df = n -1 = 5-1 = 4, we refer to the T table as follows

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For significance level 0.10, the T critical value = -2.132 and +2.132

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Inference: 2.34 does not lie in the range -2.132 and +2.132 , hence we REJECT the null hypothesis

Thus we conclude that the mean years is NOT 16

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For the significance level 0.05, the T critical values are : -2.776 and +2.776

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inference: The T statistic value -2.34 lies in the range -2.776 and +2.776, and hence we FAIL to reject the null

Conclusion; thus we conclude that the mean number of years of education is 16

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Part C)

The formula of 95% confidence interval is:

x bar - t * s/ sqrt(n) , x bar + t * s/ sqrt(n)

t critical for 95% is 2.776

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On plugging the values we get:

13.2 -2.776 *2.68 /sqrt(5) , 13.2 +2.776 *2.68/sqrt(5)

9.8729, 16.5271

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Interpretation: this implies when several samples would be taken, 95% of them will contain the true population mean number fo educated years in the range 9.8729 to 16.5271