Question

In a random sample of 100 American adults, 59% are found to be overweight. (a) (+1) Construct a 95% confidence interval for the share of American adults that are overweight.

A) Construct a 95% confidence interval for the share of American adults that are overweight.

B) A new article reports that 67% of adults in the U.S. are overweight. Is there sufficient evidence to conclude that the fraction of overweight adults in the population is below .67? Test at the 5% level.

Answer 1

a)

95% confidence interval for p is

- z * sqrt( ( 1 - ) / n) < p < + z * sqrt( ( 1 - ) / n)

0.59 - 1.96 * sqrt( 0.59 * 0.41 / 100) < p < 0.59 + 1.96 * sqrt( 0.59 * 0.41 / 100)

0.494 < p < 0.686

95% CI is ( 0.494 , 0.686 )

b)

H0 : p = 0.67

Ha: p < 0.67

Test statistics

z = - p / sqrt( p (1 - p) / n)

= 0.59 - 0.67 / sqrt( 0.67 * 0.33 / 100)

= -1.70

This is test statistics value.

p-value = P( Z < z)

= P( Z < -1.70)

= 0.0446

Since p-value < 0.05 level, we have sufficient evidence to reject H0.

We conclude at 0.05 level that we have enough evidence to support the claim.