In May 2012 President Obama made history by revealing his support of same sex ma

Question

In May 2012 President Obama made history by revealing his support of same sex marriage. Around that time the Gallup Organization polled 1,024 U.S. adults about their opinions on gay/lesbian relations and same sex marriage. They found that 54% of those sampled viewed gay/lesbian relations as "morally acceptable".

a.) Construct a 90% confidence interval for the proportion of U.S. adults who find gay/lesbian relations to be "morally acceptable". Round the margin of error to three decimal places.

b.) Does this sample provide evidence that the majority of U.S. adults (i.e., more than half) believe that gay/lesbian relations are "morally acceptable"? Use a 5% significance level. Verify that the sample size is large enough to use the normal distribution to compute the p-value for this test and include all of the details of the test.

Explanation / Answer

Part a

We are given, n = 1024, P = 0.54

Confidence level = 90%

Critical Z value = 1.6449

Confidence interval = P -/+ Z*sqrt[P*(1 – P)/n]

Standard error = SE = sqrt[P*(1 – P)/n] = sqrt(0.54*(1 – 0.54)/1024)

Standard error = sqrt(0.54*0.46/1024) = 0.015575

Margin of error = Z*Standard error

Margin of error = E = 1.6449*0.015575 = 0.026

Lower limit = P - Z*sqrt[P*(1 – P)/n]

Lower limit = P – E = 0.54 – 0.026 = 0.514

Upper limit = P + E = 0.54 + 0.026 = 0.566

Confidence interval = (0.514, 0.566)

Part b

First of all we have to check two conditions

n*p >5 and n*q > 5, where q=1 – p = 1 – 0.54 = 0.46

We are given n = 1024

n*p = 1024*0.54 = 552.96 > 5 and n*q = 1024*0.46 = 471.04 > 5

Both conditions are satisfied.

Now, we have to use one sample z test for population proportion.

Null hypothesis: H0: Half or less than half of U.S. adults believe that gay/lesbian relations are morally acceptable.

Alternative hypothesis: Ha: More than half of U.S. adults believe that gay/lesbian relations are morally acceptable.

H0: p 0.5 Versus Ha: p > 0.5

This is a one tailed test. This is a right tailed test or upper tailed test.

Level of significance = = 0.05

Test statistic = Z = (P – p) / sqrt[p*(1 – p)/n]

Where, P is sample proportion and p is population proportion.

P = 0.54, p = 0.50,

SE = sqrt[p*(1 – p)/n] = sqrt(0.54*(1 – 0.54)/1024) = 0.015575

Test statistic = Z = (0.54 – 0.50) / sqrt(0.54*(1 – 0.54)/1024)

Z = 2.568232

P-value = 0.0104

P-value < = 0.05

So, we reject the null hypothesis that Half or less than half of U.S. adults believe that gay/lesbian relations are morally acceptable.

There is sufficient evidence to conclude that more than half of U.S. adults believe that gay/lesbian relations are morally acceptable.

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