Question

1. A random sample of size n is drawn from a population that is normally distributed with a standard deviation of 8. The sample mean is found to be 50. 1.a) Construct a 98% confidence interval (CI) for the population mean uif the sample size is 16. The critical value used is The (margin of) error for the 98% confidence interval (C.I.) is The resulting Cl is 1.b) Construct a 95% confidence interval for the population mean u if the sample size is 16. 9 LThe critical value used is a arc The (margin of) error for the 95% confidence interval (C.I.) is The resulting CI is 1.c) Construct a 95 % confidence intervalfor the population mean u if the sample size is 25 The (margin of) error for the 95 % CI is The resulting CI is 1.d) How does dece as ing the level of confidence affect the (margin of) error? 1e) How does deceasing sample size affect the (margin of) error? 2. A researcher wanted to determine the mean number of hours perweeka typical person watches television. Results for the Sullivan Statistics Survey indicated that the standard deviation is 7.5 hours. 2.a) How many people are needed to estimate the number of hours people watch television perweek within 2 hours, with 95 % confidence? 2.b) How many people are needed to estimate the number of hours people watch te levision perweek within 1 hours, with 95 % confidence? es is 7. curs lonprveok t 2.c) What effect does doubling the required accuracy have on the sample size?

Answer 1

1)a) At 98% confidence level, the critical value is z_0.01 = 2.33

Margin of error = z_0.01 * $\sigma/\sqrt n$

                            = 2.33 * 8/$\sqrt {16}$

                            = 4.66

The 98% confidence interval is

$\bar x$ +/- E

= 50 +/- 4.66

= 45.34, 54.66

b) At 95% confidence level, the critical value is z_0.025 = 1.96

Margin of error = z_0.025 * $\sigma/\sqrt n$

                            = 1.96 * 8/$\sqrt {16}$

                            = 3.92

The 95% confidence interval is

$\bar x$ +/- E

= 50 +/- 3.92

= 46.08, 53.92

c) At 95% confidence level, the critical value is z_0.025 = 1.96

Margin of error = z_0.025 * $\sigma/\sqrt n$

                            = 1.96 * 8/$\sqrt {25}$

                            = 3.136

The 95% confidence interval is

$\bar x$ +/- E

= 50 +/- 3.136

= 46.864, 53.136

d) As the confidence level decreases, the margin of error also decreases.

e) As the sample size decreases, the margin of error increases.

2)a) At 95% confidence level, the critical value is z_0.025 = 1.96

Margin of error = 2

or, z_0.025 * $\sigma/\sqrt n$ = 2

or, 1.96 * 8/$\sqrt n$ = 2

or, n = (1.96 * 8/2)^2

or, n = 62

b) Margin of error = 1

or, z_0.025 * $\sigma/\sqrt n$ = 1

or, 1.96 * 8/$\sqrt n$ = 1

or, n = (1.96 * 8/1)^2

or, n = 246

c) The sample size decreases by doubling the required accuracy .