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A new brand of breakfast cereal is being market tested. One hundred boxes of the cereal were given to consumers to try. The consumers were asked whether they liked or disliked the cereal. You are given their responses below. Response Frequency Liked 60 Disliked 40 100 a. What is the point estimate of the proportion of people who will like the cereal? b. Construct a 95% confidence interval for the proportion of all consumers who will like the cereal. c. What is the margin of error for the 95% confidence interval that you constructed in part b? d. With a .95 probability, how large of a sample needs to be taken to provide a margin of error of .09 or less?

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Answer:

Check the explanation

Step-by-step explanation:

a.

point estimate of the proportion who like = 60/100 = 0.60

b)

Confidence Interval For Proportion

CI = p ± Z a/2 Sqrt(p*(1-p)/n)))

x = Mean

n = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Mean(x)=60

Sample Size(n)=100

Sample proportion = x/n =0.6

Confidence Interval = [ 0.6 ±Z a/2 ( Sqrt ( 0.6*0.4) /100)]

= [ 0.6 - 1.96* Sqrt(0.0024) , 0.6 + 1.96* Sqrt(0.0024) ]

= [ 0.504,0.696]

c)

Margin of Error = Z a/2 Sqrt(p*(1-p)/n))

x = Mean

n = Sample Size

a = 1 - (Confidence Level/100)

Za/2 = Z-table value

CI = Confidence Interval

Margin of Error = Z a/2 * ( Sqrt ( (0.6*0.4) /100) )

= 1.96* Sqrt(0.0024)

=0.096

d)

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.05 is = 1.96

Sample Proportion = 0.06

ME = 0.09

n = ( 1.96 / 0.09 )^2 * 0.06*0.94

= 26.7489 ~ 27

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