Question

An article regarding interracial dating and marriage recently appeared in a newspaper. Of the 1703 randomly selected adults, 312 identified themselves as Latinos, 322 identified themselves as blacks, 251 identified themselves as Asians, and 777 identified themselves as whites. In this survey, 86% of blacks said that their families would welcome a white person into their families.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

1.

Which distribution should you use for this problem? (Round your answers to four decimal places.)

P' ~n(.86,__)

2.Construct a 95% confidence interval for the population proportion of blacks that would welcome a white person into their family.

(i) State the confidence interval. (Round your answers to four decimal places.)

A university conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.7% of the 450 members of a fitness association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.

NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

1.Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight-year period.

(i) State the confidence interval. (Round your answers to four decimal places.)

  

Answer 1

Answering the first question

P' ~n(.86,__)

The blank should be replaced by:

n * p * (1-p)

= 322 * 0.86 * 0.14

= 38.7688

Confidence interval:

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information about the sample proportion 0.86 Sample Proportion p- 322 The critical value for a-0.05 is e1.96. The corresponding confidence interval is computed as shown below: CI(Proportion) -P e 0.86 - 196 x0.86(1-0.86 0.86 + 1.960.86(1- 0.86 322 322 (0.822,0.898) Therefore. based on the data provided, the 95% confidence interval for the population proportion is 0.822 < p < 0.898, which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.822,0.898)