Question

3. An IQ test is designed so that the mean is 100 and the standard deviation is 19 for the population of adults. Find the sample size necessary to estimate the mean IQ score of nurses such that it can be said with 99 % confidence that the sample mean is within 5 IQ points of the true mean. Assume that o = 19 and determine the required sample size using technology. The required sample size is _ (Round up to the nearest integer.) 4. In a clinical trial of a certain drug. 25 subjects experience headaches among the 220 subjects treated with the drug. Construct a 95% confidence interval estimate for the proportion of treated subjects who experience headaches, completing parts (a) through (d) below. a. Find the best point estimate of the population proportion. (Give an integer or a decimal rounded to three decimal places as needed.) p = b. Give the critical value for a = 0.05. (Round to three decimal places as needed.) Z = c. Identify the value of the margin of error for a 95% confidence interval (Give an integer or a decimal rounded to three decimal places as needed.) E = d. Construct a 95% confidence interval. (Round to three decimal places as needed.) — <p <

Answer 1

Solution :

Given that,

3) Population standard deviation = $\sigma$ = 19

Margin of error = E = 5

At 99% confidence level the z is,

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z_$\alpha$/2 = 2.576

sample size = n = [Z_$\alpha$/2* $\sigma$ / E] ²

n = [2.576 * 19 / 5 ]²

n = 95.82

Sample size = n = 96

a) Given that,

n = 220

x = 25

a) Point estimate = sample proportion = $\hat p$ = x / n = 25 / 220 = 0.114

1 - $\hat p$ = 1 - 0.114 = 0.886

b) At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

c) Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.114 * 0.886) / 220 )

= 0.042

d) A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.114 - 0.042 < p < 0.114 + 0.042

0.072 < p < 0.156