Reading 10: Sampling and Estimation-LOS j 习题精选 2011, interpret, normal, center

1215

Session 3: Quantitative Methods: Application
Reading 10: Sampling and Estimation

LOS j: Calculate and interpret a confidence interval for a population mean, given a normal distribution with 1) a known population variance, 2) an unknown population variance, or 3) an unknown variance and a large sample size.

 

 

In which one of the following cases is the t-statistic the appropriate one to use in the construction of a confidence interval for the population mean?

A) The distribution is normal, the population variance is known, and the sample size is less than 30.

B) The distribution is nonnormal, the population variance is known, and the sample size is at least 30.

C) The distribution is nonnormal, the population variance is unknown, and the sample size is at least 30.

---

 

The t-distribution is the theoretically correct distribution to use when constructing a confidence interval for the mean when the distribution is nonnormal and the population variance is unknown but the sample size is at least 30.

1215

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) $172,514 to $227,486.

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

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

---

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.

1215

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.

---

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

1215

From a sample of 41 monthly observations of the S& Mid-Cap index, the mean monthly return is 1% and the sample variance is 36. For which of the following intervals can one be closest to 95% confident that the population mean is contained in that interval?

A) 1.0% ± 1.9%.

B) 1.0% ± 6.0%.

C) 1.0% ± 1.6%.

---

If the distribution of the population is nonnormal, but we don’t know the population variance, we can use the Student’s t-distribution to construct a confidence interval. The sample standard deviation is the square root of the variance, or 6%. Because there are 41 observations, the degrees of freedom are 40. From the Student’s t distribution, we can determine that the reliability factor for t_0.025, is 2.021. Then the 95% confidence interval is 1.0% ± 2.021(6 / √41) or 1.0% ± 1.9%.

1215

A random sample of 25 Indiana farms had a mean number of cattle per farm of 27 with a sample standard deviation of five. Assuming the population is normally distributed, what would be the 95% confidence interval for the number of cattle per farm?

A) 23 to 31.

B) 25 to 29.

C) 22 to 32.

---

The standard error of the sample mean = 5 / √25 = 1
Degrees of freedom = 25 ? 1 = 24
From the student’s T table, t_5/2 = 2.064
The confidence interval is: 27 ± 2.064(1) = 24.94 to 29.06 or 25 to 29.

1215

A local high school basketball team had 18 home games this season and averaged 58 points per game. If we assume that the number of points made in home games is normally distributed, which of the following is most likely the range of points for a confidence interval of 90%?

A) 34 to 82.

B) 24 to 78.

C) 26 to 80.

---

This question has a bit of a trick. To answer this question, remember that the mean is at the midpoint of the confidence interval. The correct confidence interval will have a midpoint of 58. (34 + 82) / 2 = 58.

1215

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) 2.313 to 2.687.

B) 2.387 to 2.613.

C) 2.410 to 2.589.

---

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.)

1215

A sample of 100 individual investors has a mean portfolio value of $28,000 with a standard deviation of $4,250. The 95% confidence interval for the population mean is closest to:

A) $19,500 to $28,333.

B) $27,159 to $28,842.

C) $27,575 to $28,425.

---

Confidence interval = mean ± t_c{S / √n}

= 28,000 ± (1.98) (4,250 / √100) or 27,159 to 28,842

If you use a z-statistic because of the large sample size, you get 28,000 ± (1.96) (4,250 / √100) = 27,167 to 28,833, which is closest to the correct answer.

1215

The average return on small stocks over the period 1926-1997 was 17.7%, and the standard error of the sample was 33.9%. The 95% confidence interval for the return on small stocks in any given year is:

A) 16.8% to 18.6%.

B) –16.2% to 51.6%.

C) –48.7% to 84.1%.

---

A 95% confidence level is 1.96 standard deviations from the mean, so 0.177 ± 1.96(0.339) = (–48.7%, 84.1%).

1215

Books Fast, Inc., prides itself on shipping customer orders quickly. Downs Shipping Service has promised that mean delivery time will be less than 72 hours. Books Fast sampled 27 of its customers and found a mean delivery time of 76 hours, with a sample standard deviation of 6 hours. Based on this sample and assuming a normal distribution of delivery times, what is the confidence interval at 5% significance?

A) 73.63 to 78.37 hours.

B) 68.50 to 83.50 hours.

C) 65.75 to 86.25 hours.

---

The confidence interval is equal to 76 + or ? (2.056)(6 / √27) = 73.63 to 78.37 hours.
Because the sample size is small, we use the t-distribution with (27 ? 1) degrees of freedom.