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Chapter 6: Q44E (page 356)

A random sample of 50 consumers taste-tested a new snack food. Their responses were coded (0: do not like; 1: like; 2: indifferent) and recorded as follows:

a. Use an 80% confidence interval to estimate the proportion of consumers who like the snack food.

b. Provide a statistical interpretation for the confidence interval you constructed in part a.

Short Answer

Expert verified

a. The 80% confidence interval is0.2170,0.3830

b.One is 80% confident that the true proportion of consumers who like the snack food lies between the interval0.2170,0.3830

Step by step solution

01

Given information

A random sample of 50 consumers taste-tested a new snack food. Their responses were coded according to some criteria (0: do not like; 1: like; 2: indifferent). This has been presented in the table above.

02

(a) Calculation of 80% confidence interval of estimate the proportion of consumers who like the snack food.

The sample size is n= 50

15 of the data value are a 1, which indicates x=15

p=xn=1550=0.3

Here the confidence coefficient is 80%, hence α=20%.Now from the standard normal distribution table,zα2=1.28

The margin of error

E=zα2×p^1-p^n=1.28×0.3×1-0.350=0.0830

Hence 80% lower limit:p^-E=0.3-0.0830=0.2170

80% upper limit:p^+E=0.3+0.0830=0.3830

The 80% confidence interval is0.2170,0.3830

03

(b) Interpretation of confidence interval constructed in part a

One is 80% confident that the true proportion of consumers who like snack food lies between the interval0.2170,0.3830

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