Question

* You wanted to estimate the mean number of vehicles crossing a busy bridge in your neighborhood each morning during rush hour for the past year. To accomplish this, you stationed yourself and a few assistants at one end of the bridge on 31 randomly selected mornings during the year and counted the number of vehicles crossing the bridge in a 10-minute period during rush hour. You found the mean to be 119 vehicles per minute, with a standard deviation of 31.

(a) Construct the 95% confidence interval for the population mean (vehicles per minute).

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(b) Construct the 99% confidence interval for the population mean (vehicles per minute)

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* A college counselor wants to determine the average amount of time first-year students spend studying. He randomly samples 31 students from the freshman class and asks them how many hours a week they study. The mean of the resulting scores is 17 hours, and the standard deviation is 5.8 hours.

(a) Construct the 95% confidence interval for the population mean.
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(b) Construct the 99% confidence interval for the population mean.

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*A cognitive psychologist believes that a particular drug improves short-term memory. The drug is safe, with no side effects. An experiment is conducted in which 8 randomly selected subjects are given the drug and then given a short time to memorize a list of 10 words. The subjects are then tested for retention 15 minutes after the memorization period. The number of words correctly recalled by each subject is as follows: 6, 11, 12, 4, 6, 7, 8, 6. Over the past few years, the psychologist has collected a lot of data using this task with similar subjects. Although he has lost the original data, he remembers that the mean was 4.9 words correctly recalled and that the data were normally distributed.

(a) On the basis of these data, what can we conclude about the effect of the drug on short-term memory? Use α = 0.05_{2 tail} in making your decision.

tobt =

t_crit = ±

Answer 1

Question 1

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 31- 1 ) = 2.042
119 ± t(0.05/2, 31 -1) * 31/√(31)
Lower Limit = 119 - t(0.05/2, 31 -1) 31/√(31)
Lower Limit = 107.6306
Upper Limit = 119 + t(0.05/2, 31 -1) 31/√(31)
Upper Limit = 130.3694
95% Confidence interval is ( 107.6306 , 130.3694 )

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 31- 1 ) = 2.75
119 ± t(0.01/2, 31 -1) * 31/√(31)
Lower Limit = 119 - t(0.01/2, 31 -1) 31/√(31)
Lower Limit = 103.6886
Upper Limit = 119 + t(0.01/2, 31 -1) 31/√(31)
Upper Limit = 134.3114
99% Confidence interval is ( 103.6886 , 134.3114 )

Question 2

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 31- 1 ) = 2.042
17 ± t(0.05/2, 31 -1) * 5.8/√(31)
Lower Limit = 17 - t(0.05/2, 31 -1) 5.8/√(31)
Lower Limit = 14.8728
Upper Limit = 17 + t(0.05/2, 31 -1) 5.8/√(31)
Upper Limit = 19.1272
95% Confidence interval is ( 14.8728 , 19.1272 )

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 31- 1 ) = 2.75
17 ± t(0.01/2, 31 -1) * 5.8/√(31)
Lower Limit = 17 - t(0.01/2, 31 -1) 5.8/√(31)
Lower Limit = 14.1353
Upper Limit = 17 + t(0.01/2, 31 -1) 5.8/√(31)
Upper Limit = 19.8647
99% Confidence interval is ( 14.1353 , 19.8647 )