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1)Suppose the dean believes that the average salary of the population should be about $58k per year, with a standard deviation of $2k
1)Suppose the dean believes that the average salary of the population should be about $58k per year, with a standard deviation of $2k. We need to conclude that the mean salary is less than what the dean has believed to be (30points, Review PPT Chapter 10). (Hint: Hypothesis Testing and Confidence Interval - One sample; population standard deviation is known)
- What are the null and alternate hypotheses ?
- What is the level of significance ?
- What is the standard error ? 2000/(121^0.5) =
- Decide on the test statistic and calculate the value of the test statistic (hint: write the equation and calculate the statistic) ?
- What’s your decision regarding the hypothesis and interpret the result using test-score rejection region rule or p value rule .
- Suppose the population standard deviation of $2k (20points, Review PPT Chapter 9): (Hint: Hypothesis Testing and Confidence Interval - One sample; population standard deviation is known)
- What is the Z critical value of 95% confidence interval ?
- What is the standard error ? 2000/(121^0.5) =
- Please estimate the confidence interval of the population’s average salary (hint: write down the equation and calculate the results)
60,000 +/- 1.96(181.818)
60,000 +/- 356.36
- Suppose the standard deviation of the collected data from 121 alumni is $3k. Can we conclude that the true mean of alumni salary is different from $58k (20 points; Review PPT Chapter 10)? (Hint: Hypothesis Testing and Confidence Interval - One sample; sample standard deviation is known)
- What are the null and alternate hypotheses ?
- Decide on the test statistic and calculate the value of the test statistic (hint: write the equation and calculate the statistic.
t = (60,000-58,000) / 3000/(121^0.5)
- Please make the conclusion .
- Suppose that the standard deviation of the collected data from 121 alumni is $3k and the mean is $60k, please estimate the confidence interval of the population’s average salary (10 points, Review PPT Chapter 9). (Hint: Hypothesis Testing and Confidence Interval - One sample; sample standard deviation is known)
95% t = 121-1
t = 1.980
60,000 +/- 1.980(3000/11)
60,000 +/- 540
- The dean wants to know whether there is salary difference between male and female alumni. He knows that the population standard deviation for male is $3 and for female is $4. The sampled data show that the average salary for 100 male is $60k and the average salary for 144 female is $56k (20 points, Review PPT Chapter 11). (Hint: Hypothesis Testing and Confidence Interval - two samples; population standard deviation is known)
- What are the null and alternate hypotheses ?
Rule: Reject H0 if z < -1.645 or > 1.645
- Decide on the test statistic and calculate the value of the test statistic (hint: write the equation and calculate the statistic) ?
Z = 60,000-56,000 / (9/100 + 16/144)^0.5
Z = 4,000 / .44845
- What’s your decision regarding the hypothesis and interpret the result using test-score rejection region rule or p value rule .
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