Question

In: Statistics and Probability

A random sample of 28 items is drawn from a population whose standard deviation is unknown....

A random sample of 28 items is drawn from a population whose standard deviation is unknown. The sample mean is x⎯⎯x¯ = 790 and the sample standard deviation is s = 15. Use Appendix D to find the values of Student’s t.

(a)

Construct an interval estimate of μ with 99% confidence. (Round your answers to 3 decimal places.)

  The 99% confidence interval is from  to
(b)

Construct an interval estimate of μ with 99% confidence, assuming that s = 30. (Round your answers to 3 decimal places.)

  The 99% confidence interval is from  to
(c)

Construct an interval estimate of μ with 99% confidence, assuming that s = 60. (Round your answers to 3 decimal places.)

  The 99% confidence interval is from  to
(d)

Describe how the confidence interval changes as s increases.

The interval stays the same as s increases.
The interval gets wider as s increases.
The interval gets narrower as s increases.

Solutions

Expert Solution

a)

t critical value at 0.01 significance level with 27 df = 2.771

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

790 - 2.771 * 15 / sqrt(28) < < 790 + 2.771 * 15 / sqrt(28)

782.145 < < 797.855

99% CI is from 782.145 to 797.855

b)

t critical value at 0.01 significance level with 27 df = 2.771

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

790 - 2.771 * 30 / sqrt(28) < < 790 + 2.771 * 30 / sqrt(28)

774.290 < < 805.710

99% CI is from 774.290  to 805.710

c)

t critical value at 0.01 significance level with 27 df = 2.771

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

790 - 2.771 * 60 / sqrt(28) < < 790 + 2.771 * 60 / sqrt(28)

758.580 < < 821.420

99% CI is from 758.580 to 821.420

d)

The interval gets wider as S increases.


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