honeycfa

A traffic engineer is trying to measure the effects of carpool-only lanes on the expressway. Based on a sample of 20 cars at rush hour, he finds that the mean number of occupants per car is 2.5, with a standard deviation of 0.4. If the population is normally distributed, what is the confidence interval at the 5% significance level for the number of occupants per car?

A) 2.387 to 2.613.

B) 2.410 to 2.589.

C) 2.313 to 2.687.

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The reliability factor corresponding with a 5% significance level (95% confidence level) for the Student’s t-distribution with (20 ? 1) degrees of freedom is 2.093. The confidence interval is equal to: 2.5 ± 2.093(0.4 / √20) = 2.313 to 2.687. (We must use the Student’s t-distribution and reliability factors because of the small sample size.)

honeycfa

The average salary for a sample of 61 CFA charterholders with 10 years experience is $200,000, and the sample standard deviation is $80,000. Assume the population is normally distributed. Which of the following is a 99% confidence interval for the population mean salary of CFA charterholders with 10 years of experience?

A) $172,754 to $227,246.

B) $172,514 to $227,486.

C) $160,000 to $240,000.

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If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 61 observations, the degrees of freedom are 60. From the student’s t table, we can determine that the reliability factor for t_α/2, or t_0.005, is 2.660. Then the 99% confidence interval is $200,000 ± 2.660($80,000 / √61) or $200,000 ± 2.660 × $10,243, or $200,000 ± $27,246.

honeycfa

From a sample of 41 orders for an on-line bookseller, the average order size is $75, and the sample standard deviation is $18. Assume the distribution of orders is normal. For which interval can one be exactly 90% confident that the population mean is contained in that interval?

A) $71.29 to 78.71.

B) $70.27 to $79.73.

C) $74.24 to $75.76.

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If the distribution of the population is normal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. Because there are 41 observations, the degrees of freedom are 40. From the student’s t table, we can determine that the reliability factor for t_α/2, or t_0.05, is 1.684. Then the 90% confidence interval is $75.00 ± 1.684($18.00 / √41), or $75.00 ± 1.684 × $2.81 or $75.00 ± $4.73