Question

7. Suppose that a particular brand of peanut butter is in fact favored by 47% of all moms. A consumer organization will take a random sample of 200 moms from across the country and will use ˆp, the sample proportion, to estimate p, the actual proportion of moms who favor that peanut butter brand.

(a) Is it reasonable to assume that the sampling distribution of ˆp is approximately normal? Justify.

(b) Compute the mean and standard deviation of the ˆp distribution. (Give your answer to four decimal places.)

(c) What is the approximate probability that the random sample will produce a ˆp value greater than .5, causing the consumer organization to incorrectly conclude that the majority of moms prefer this brand of peanut butter? (Give your answer to four decimal places.)

Answer 1

Solution

Given that,

p = 0.47

1 - p = 1 - 0.47 = 0.53

n = 200

a) The sampling distribution of is approximately normal, because

np 10 and n(1 - p)

200* 0.47 = 94 and 200 * 0.53 = 106 10

b)   = p = 0.47

=  [p( 1 - p ) / n] = [(0.47 * 0.53) / 200 ] = 0.0353

c) P( > 0.50) = 1 - P( < 0.50 )

= 1 - P(( - ) / < (0.50 - 0.47) / 0.0353)

= 1 - P(z < 0.85 )

Using z table

= 1 - 0.8023

= 0.1977