It has been reported that 10.3% of U.S. households do not own a vehicle, with 34.2% owning 2 vehicles, and 17.1% owning 3 or more vehicles. The data for a random sample of 100 households in a resort community are summarized in the frequency distribution below. At the 0.05 level of significance, can we reject the possibility that the vehicle-ownership distribution in this community differs from that of the nation as a whole?
Number of Numberof
Vehicles Owned Households
0 20
1 35
2 23
3 or more 22
100
Since this is the fourth time I'm reading the same thing, I suspect you WERE trying to cut and paste.
Sra
To determine whether the vehicle-ownership distribution in the resort community differs from that of the nation as a whole, we can perform a chi-square goodness-of-fit test. This test compares the observed frequencies in each category (number of vehicles owned) with the expected frequencies based on the national distribution.
Here are the steps to conduct the chi-square test:
1. State the null and alternative hypotheses:
- Null hypothesis (H0): The vehicle-ownership distribution in the resort community is the same as that of the nation as a whole.
- Alternative hypothesis (Ha): The vehicle-ownership distribution in the resort community differs from that of the nation as a whole.
2. Calculate the expected frequencies based on the national distribution:
- We can calculate the expected frequencies for each category by multiplying the respective national percentages by the total number of households in the resort community (100).
Expected frequency of households with 0 vehicles: 0.103 * 100 = 10.3
Expected frequency of households with 1 vehicle: 0.448 * 100 = 44.8
Expected frequency of households with 2 vehicles: 0.342 * 100 = 34.2
Expected frequency of households with 3 or more vehicles: 0.171 * 100 = 17.1
3. Calculate the test statistic:
- The chi-square test statistic can be calculated using the formula:
X^2 = Σ((Observed frequency - Expected frequency)^2 / Expected frequency)
For each category, calculate (Observed frequency - Expected frequency)^2 / Expected frequency and sum them up.
Calculating for the given data:
(20 - 10.3)^2 / 10.3 = 8.573
(35 - 44.8)^2 / 44.8 = 4.526
(23 - 34.2)^2 / 34.2 = 3.333
(22 - 17.1)^2 / 17.1 = 1.134
Summing up these values: 8.573 + 4.526 + 3.333 + 1.134 = 17.566
4. Determine the degrees of freedom (df):
- Degrees of freedom = Number of categories - 1
We have four categories, so df = 4 - 1 = 3
5. Look up the critical value:
- At a significance level of 0.05 and df = 3, the critical value of the chi-square distribution is 7.815.
6. Compare the test statistic and critical value:
- If the test statistic is greater than the critical value, we can reject the null hypothesis.
In this case, the test statistic (17.566) is greater than the critical value (7.815), so we reject the null hypothesis.
7. Interpretation:
- Based on the data, we have evidence to suggest that the vehicle-ownership distribution in the resort community differs from that of the nation as a whole at the 0.05 level of significance.