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SCI1020 - Introduction to Statistical Reasoning - S2 2025

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 A random group of eighty-five students in Melbourne high schools takes a course designed to improve their ATAR scores. Based on these students, a 90% confidence interval for the mean improvement in ATAR scores from this course for all Melbourne high school students is computed as (72.3, 91.4) points. The correct interpretation of this interval is:

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The amount of time customers at the “Speedy” tyre store spend waiting for their car tyres to be checked has a (unknown) mean \mu\mu and standard deviation \sigma\sigma = 4 minutes. It is company policy that the customer wait time should be 20 minutes (or less). The manager of the store selects a random sample of 150 customer wait times and observes a mean wait time of 21 minutes.

A 99% confidence interval for the population mean wait time based on this sample was calculated to be (20.16, 21.84) minutes. 

Which of the following statements is true?

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The amount of time customers at a “Quick-Change” motor oil store spend waiting for their cars to be serviced has the Normal distribution with mean μ and standard deviation σ = 4 minutes. It is company policy that the customer wait time should be 20 minutes (or less). The manager of a particular store selects a random sample of 150 customer wait times and observes a mean wait time of 21 minutes.

A 99% confidence interval for the population mean wait time μ based on this sample was calculated to be (20.16, 21.84) minutes.

Suppose that the population standard deviation σ = 4 was wrong and the correct population standard deviation should be σ =5, a new 99% confidence interval for the same sample was calculated using σ = 5. The new 99% confidence interval

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You plan to construct a 95% confidence interval for the mean of a Normal population with (known) standard deviation . Which of the following can you change in practice to reduce the size of the margin of error?

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A researcher used a new drug to treat 100 subjects with high cholesterol. For the patients in the study, after two months of treatment the average decrease in cholesterol level was 80 milligrams per decilitre (mg/dl). Assume that the decrease in cholesterol after two months of taking the drug follows a Normal distribution, with unknown mean μ and standard deviation σ = 20 mg/dl. The researcher will construct a 90% confidence interval to estimate μ. What is the margin of error?

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I computed a 95% confidence interval for the mean lifetime of a set of tires as (37,000, 42,000) kilometres. Based on this interval, I know:

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The critical value, z*, used for constructing a 98% confidence interval for a population mean μ is:

[read from the appropriate statistical table in the textbook]

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The time (in number of days) until maturity of a certain variety of tomato plant is Normally distributed with unknown mean μ and standard deviation σ = 2.4. A simple random sample of four plants of this variety is selected. The number of days until maturity for each plant is given below

63     69     62     66

Based on these data, a 99% confidence interval for μ, in days, is

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A researcher is interested in the average time served in jail for robbery. They take a sample of 400 convictions, and find the average time served is

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= 7.5 years, with standard deviation s = 3 years. Since the sample is so large, it is reasonable to believe that s is close to σ. A 95% confidence interval for the average time served in jail for robbery is:

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The amount of time customers at a “Quick-Change” motor oil store spend waiting for their cars to be serviced has the Normal distribution with mean μ and standard deviation σ = 4 minutes. It is company policy that the customer wait time should be 20 minutes (or less). The manager of a particular store selects a random sample of 150 customer wait times and observes a mean wait time of 21 minutes.

A 95% confidence interval for the mean waiting time for customers is about

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