To test the newspaper's claim, we will use a two-proportion z-test.
The formula for the z-test for two proportions is:
z = (p1 - p2) / √(p * (1-p) * (1/n1 + 1/n2))
Where:
p1 = proportion of singles purchasing houses now
p2 = proportion of singles purchasing houses 5 years ago
p = pooled proportion = (x1 + x2) / (n1 + n2)
x1 = number of singles purchasing houses now
x2 = number of singles purchasing houses 5 years ago
n1 = total number of house purchases now
n2 = total number of house purchases 5 years ago
Given data:
x1 = 150 (number of singles purchasing houses now)
x2 = 129 (number of singles purchasing houses 5 years ago)
n1 = 500 (total number of house purchases now)
n2 = 500 (total number of house purchases 5 years ago)
Calculations:
p1 = x1 / n1 = 150 / 500 = 0.3
p2 = x2 / n2 = 129 / 500 = 0.258
p = (x1 + x2) / (n1 + n2) = (150 + 129) / (500 + 500) = 0.279
Now we can plug these values into the formula to calculate the test statistic:
z = (p1 - p2) / √(p * (1-p) * (1/n1 + 1/n2))
= (0.3 - 0.258) / √(0.279 * (1-0.279) * (1/500 + 1/500))
= 0.042 / √(0.078804 * 0.721196 * 0.004)
= 0.042 / √(0.000228290076)
= 0.042 / 0.015108724
= 2.778151
Rounding to two decimal places, the test statistic is approximately 2.78.
A newspaper story claims that more houses are purchased by singles now than singles 5 years ago. To test this claim, two studies were conducted on the buying habits of singles over the past 5 years. In the first study, 500 house purchases in the current year were randomly selected and 150 of those were made by singles. In the second study, again 500 house purchases were randomly selected from 5 years ago and 129 of those were made by single people. Test the newspaper’s claim using a 0.05 level of significance. Is there sufficient evidence to support the newspaper’s claim? Let singles now be Population 1 and let singles 5 years ago be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
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