This Question: 1 pt 1 of 4 (0 complete) This Quiz: 4 pts possible A producer of

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This Question: 1 pt 1 of 4 (0 complete) This Quiz: 4 pts possible A producer of a variety of salty snacks would like to estimate the average weight of a bag of BBQ potato chips produced during the filling process at one of its plants. Determine the sample size needed to construct a 99% confidence interval with a margin of error equal to 0.005 ounces. Assume the standard deviation for the potato chip filling process is 0.02 ounces The sample size needed is (Round up to the nearest integer.) Enter your answer in the answer box.

Explanation / Answer

Solution:

1) For 99%, z = 2.576
d = 0.005, sd = 0.02

Sample size = (z2*s2)/d2

= (2.5762*0.022)/0.0052

= 106.17

The sample size needed is = 107

2)

a. 90% Confidence interval for p
p = x/n = 91/400 = 0.2275
Significance level = 1-confidence = 1-0.90 = 0.10
Critical z-value = Z/2 = Z0.1/2 = Z0.05 = 1.645

Standard error of p : SE = sqrt(p *(1-p)/n)
= sqrt((0.2275 * 0.7725)/400)
= 0.020960
b. Margin of Error E = Z/2 * SE
= 1.645 * 0.020960
= 0.0344792
90% Confidence Interval is given by
= 0.2275 ± 0.0344792
= (0.1930208, 0.2619792)
c. From previous history it is known that the proportion of adult drivers who admittied to texting while driving is 0.32. From part a), we observe that 90% confidence interval based on the sample observation do not contain this value of proportion. Thus, the sample provides a strong evidence to conclude that the proportion has reduced from 0.32.

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