Question

The CDC periodically administers large randomized surveys to track the health of Americans.
In a survey of 4378 adults ending in 2016, 70.9% were overweight, and
in a survey of 4246 adults ending in 2012, 68.7% were overweight.

(a) The standard error for estimating the change in population proportion is 0.009888.
Calculate the 95% confidence interval for the change in population proportion
to 4 decimal accuracy: [ x.xxxx, x.xxxx ]

(b) Test the hypothesis that the population proportions are equal vs the two-sided alternative hypothesis.
Calculation accuracy required:
Test statistic: 4 decimals x.xxxx
p-value: 4 decimals 0.xxxx

Answer 1

(a) The z critical = 1.96

the difference in proportions = 0.709 - 0.687 = 0.022

The lower Limit = 0.022 - (1.96 * 0.009888) = 0.022 - 0.0194 = 0.0026

The Upper Limit = 0.022 + (1.96 * 0.009888) = 0.022 + 0.0194 = 0.0414

The 95% CI is [0.0026, 0.0414]

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(b) p1 = 0.709, n1 = 4378, x1 = 0.709 * 4378 = 3104

p2 = 0.687, n2 = 4246, x2 = 0.687 * 4246 = 2917

Therefore pooled proportion = p = (3104 + 2917) / (4378 + 4246) = 6021/8624 = 0.6982

1 - p = 0.3018

= 0.05 (default level)

(a) The Hypothesis:

H0: p₁ = p_{2 (Claim)}

Ha: p₁ p₂

This is a 2 Tailed Test.

The Test Statistic:

The p Value:    The p value (2 Tail) for Z = 2.2251, is; p value = 0.0260

The Critical Value:   The critical value (2 tail) at = 0.05, Zcritical = +1.96 and -1.96

The Decision Rule:    If Zobserved is > Zcritical or if Zobserved is < -Zcritical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision:    Since Z observed (2.2251) is > Zcritical (1.96), We Reject H0.

Also since P value (0.0260) is < (0.05), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to warrant rejection of the claim that the two population proportions are equal.

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