New research shows that members of a certain youth generation have a great say i
ID: 3301088 • Letter: N
Question
New research shows that members of a certain youth generation have a great say in household purchases. Specifically, 61 % of the people in the generation have a say in computer purchases. Suppose you select a sample of 100 respondents from the generation. Complete parts (a) through (d) below.
a. What is the probability that the sample percentage will be contained between 55 % and 62 %? . 4739 (Type an integer or decimal rounded to four decimal places as needed.)
b. The probability is 90 % that the sample percentage will be contained within what symmetrical limits of the population percentage? The probability is 90 % that the sample percentage will be contained above nothing % and below %.
Explanation / Answer
Pr(People in this generation have a say in computer purchases) p = 0.61
Sample size = 100
standard error of the proportion se0= sqrt [p * (1-p)/ N] = sqrt [0.61 * 0.39/100] = 0.0488
(a) Pr (0.55 < p < 0.62) = Pr( p < 0.62) - Pr( p < 0.55)
= (Z2 ) - (Z1 )
Z2 = (0.62 - 0.61)/ 0.0488 = 0.205
Z1 = (0.55 - 0.61)/ 0.0488 = -1.23
Pr (0.55 < p < 0.62) = Pr( p < 0.62) - Pr( p < 0.55)
= (Z2 ) - (Z1 ) = (0.205) - (-1.23)
= 0.5813 - 0.1093 = 0.4720
(b) we have to find the sample percentage that will contain 90% of the sample percentage.
so We have to find 90% confidence interval = p +- Z90% se0
= 0.61 +- 1.645 * 0.0488
= (0.5297, 0.6903)
so the probability is 90% that the sample percentage will be contained above is 69.03% and 52.97%