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.
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.
To test the newspaper's claim, we will compute the value of the test statistic using the sample proportions from the two studies.
The sample proportion for singles purchasing houses in the current year is calculated as follows:
Sample proportion for singles in the current year (p1) = Number of house purchases by singles in the current year / Total number of house purchases in the current year
p1 = 150 / 500 = 0.30
The sample proportion for singles purchasing houses 5 years ago is calculated as follows:
Sample proportion for singles 5 years ago (p2) = Number of house purchases by singles 5 years ago / Total number of house purchases 5 years ago
p2 = 129 / 500 = 0.258
To compare the sample proportions, we will compute the test statistic called the z-score, which measures the difference between the two proportions.
The formula for the z-score is given by:
z = (p1 - p2) / √(p̂(1-p̂)(1/n1 + 1/n2))
where p̂ is the pooled sample proportion, given by:
p̂ = (x1 + x2) / (n1 + n2)
x1 and x2 are the total number of singles purchasing houses in the current year and 5 years ago respectively, and n1 and n2 are the total number of house purchases in the current year and 5 years ago respectively.
Using the given values:
n1 = 500, n2 = 500, x1 = 150, x2 = 129
Therefore,
p̂ = (150 + 129) / (500 + 500) = 279 / 1000 = 0.279
Now, we can substitute the values into the formula for the z-score:
z = (p1 - p2) / √(p̂(1-p̂)(1/n1 + 1/n2))
= (0.30 - 0.258) / √(0.279(1-0.279)(1/500 + 1/500))
≈ 0.042 / √(0.077913)(0.004)
Round the result to two decimal places:
z ≈ 0.042 / 0.004
z ≈ 10.50
The value of the test statistic (z-score) is approximately 10.50.
To test the newspaper's claim, we need to compute the test statistic. In this case, we will use a two-proportion z-test to compare the proportions of singles purchasing houses now and five years ago.
The test statistic for comparing two proportions is given by:
z = (p₁ - p₂) / sqrt( p̂ * (1 - p̂) * (1/n₁ + 1/n₂) )
Where:
p₁ = proportion of singles purchasing houses now
p₂ = proportion of singles purchasing houses five years ago
p̂ = (x₁ + x₂) / (n₁ + n₂)
x₁ = number of singles purchasing houses now
x₂ = number of singles purchasing houses five years ago
n₁ = sample size for houses purchased now
n₂ = sample size for houses purchased five years ago
In this case:
x₁ = 150 (from the first study)
x₂ = 129 (from the second study)
n₁ = n₂ = 500 (from both studies)
Let's calculate the values:
p̂ = (150 + 129) / (500 + 500) = 279 / 1000 = 0.279
Now we can substitute the values into the formula and calculate the test statistic:
z = (0.279 - p₂) / sqrt( 0.279 * (1 - 0.279) * (1/500 + 1/500) )
To calculate the value of p₂, we need to divide the number of singles purchasing houses five years ago by the total sample size for five years ago:
p₂ = 129 / 500 = 0.258
Now we can substitute the values into the formula and calculate the test statistic:
z = (0.279 - 0.258) / sqrt( 0.279 * (1 - 0.279) * (1/500 + 1/500) )
z = 0.021 / sqrt( 0.279 * 0.721 * 0.004 )
z ≈ 0.021 / 0.015
z ≈ 1.40 (rounded to two decimal places)
The value of the test statistic is approximately 1.40.