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Homework answers / question archive / 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 H_{0} 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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