Question

Question 19 (5 points) A pollster wants to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better. A Gallup poll taken in July 2010 estimates this proportion to be 0.33. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.039? Write only an integer as your answer. Your Answer: Answer Question 20 (5 points) A researcher wants to construct a 99% confidence interval for the proportion of people who changed jobs in the past year. What sample size is needed so that the confidence interval will have a margin of error of 0.078 ? Write only an integer as your answer. Your Answer: Answer

Answer 1

Solution :

Given that,

$\hat p$= 0.33

1 - $\hat p$ = 1 - 0.33 = 0.67

margin of error = E = 0.039

At 95% confidence level the z is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z$\alpha$/2 = Z0.025 = 1.96 ( Using z table ( see the 0.025 value in standard normal (z) table corresponding z value is 1.96 )

Sample size = n = (Z$\alpha$/2 / E)2 * $\hat p$ * (1 - $\hat p$ )

= (1.96 / 0.039)2 * 0.33 * 0.67

= 558.43

Sample size = 559 ROUNDED

B

Solution :

Given that,

$\hat p$ = 0.5

1 - $\hat p$ = 1 - 0.5= 0.5

margin of error = E =0.078

At 99% confidence level the z is,

$\alpha$ = 1 - 99%

$\alpha$ = 1 - 0.99 = 0.01

$\alpha$/2 = 0.005

Z$\alpha$/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )

Sample size = n = (Z$\alpha$/2 / E)2 * $\hat p$ * (1 - $\hat p$ )

= (2.58 / 0.078)2 * 0.5 * 0.5

=274

Sample size = 274