Question

People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March, 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior. Round your answers to four decimal places. a. Show the sampling distribution of p, the proportion of groceries thrown out by your sample respondents. 1. Is the distribution normal or abnormal? 2. The Mean is ? What is standard error?

b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion? c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion? In determining your answer, use the standard error found in part a. and the probability found using the tables in the text.

Answer 1

A) 1) n = 540

p = 0.12

2) Mean = $\mu_{\widehat p}$ = p = 0.12

Standard deviation = sqrt(p(1 - p)/n)

= sqrt(0.12 * (1 - 0.12)/540)

= 0.014

b) P(0.09 < $\widehat p$ < 0.15)

= P((0.09 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < ($\widehat p$ - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < (0.15 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$)

= P((0.09 - 0.12)/0.014 < Z < (0.15 - 0.12)/0.014)

= P(-2.14 < Z < 2.14)

= P(Z < 2.14) - P(Z < -2.14)

= 0.9838 - 0.0162

= 0.9676

C) P(0.105 < $\widehat p$ < 0.135)

= P((0.105 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < ($\widehat p$ - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$ < (0.135 - $\mu_{\widehat p}$ )/$\sigma_{\widehat p}$)

= P((0.105 - 0.12)/0.014 < Z < (0.135 - 0.12)/0.014)

= P(-1.07 < Z < 1.07)

= P(Z < 1.07) - P(Z < -1.07)

= 0.8577 - 0.1423

= 0.7154