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
In the future, try to post the same item only once.
Sra
To determine whether the vehicle-ownership distribution in the resort community differs significantly from that of the nation as a whole, we need to perform a chi-square goodness-of-fit test.
Step 1: State the hypotheses:
- Null Hypothesis (H0): The vehicle-ownership distribution in the resort community is the same as the distribution in the nation as a whole.
- Alternative Hypothesis (H1): The vehicle-ownership distribution in the resort community differs from the distribution in the nation as a whole.
Step 2: Set the significance level:
- The given significance level is 0.05. This means we want to be 95% confident in our results.
Step 3: Calculate the expected frequencies:
- To compare the observed frequencies (from the resort community) with the expected frequencies (based on the national distribution), we need to calculate the expected frequencies for each category.
- We do this by multiplying the proportion of each category in the national distribution by the total number of households in the resort community sample (100).
So, according to the national distribution:
- 10.3% of households do not own a vehicle, which is equivalent to 10.3/100 * 100 = 10.3 households.
- 34.2% of households own 2 vehicles, which is equivalent to 34.2/100 * 100 = 34.2 households.
- 17.1% of households own 3 or more vehicles, which is equivalent to 17.1/100 * 100 = 17.1 households.
- The remaining percentage (100 - 10.3 - 34.2 - 17.1 = 38.4%) represents households that own only 1 vehicle, which is equivalent to 38.4/100 * 100 = 38.4 households.
We can now organize the expected frequencies based on these calculations:
Number of Vehicles Owned | Observed Frequency | Expected Frequency
0 | 20 | 10.3
1 | 35 | 38.4
2 | 23 | 34.2
3 or more | 22 | 17.1
Step 4: Calculate the chi-square test statistic (X2):
- The chi-square test statistic measures the difference between the observed and expected frequencies. It determines how well the observed frequencies fit the expected frequencies based on chance alone.
- The formula to calculate the chi-square test statistic is: X2 = Σ((Observed - Expected)² / Expected)
- Calculate the value for each category, and sum them up to obtain X2.
For the "0 vehicles" category:
((20 - 10.3)² / 10.3) = 8.94
For the "1 vehicle" category:
((35 - 38.4)² / 38.4) = 0.28
For the "2 vehicles" category:
((23 - 34.2)² / 34.2) = 4.69
For the "3 or more vehicles" category:
((22 - 17.1)² / 17.1) = 1.16
Summing up these values: X2 = 8.94 + 0.28 + 4.69 + 1.16 = 15.07
Step 5: Determine the degrees of freedom:
- The degrees of freedom (df) is the number of categories minus 1. In this case, there are 4 categories, so df = 4 - 1 = 3.
Step 6: Determine the critical value:
- Look up the critical value for a chi-square test with 3 degrees of freedom at a 0.05 significance level. In this case, the critical value is 7.815.
Step 7: Compare the test statistic with the critical value:
- If the test statistic (X2) is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
- In this case, X2 (15.07) > critical value (7.815), so we reject the null hypothesis.
Step 8: State the conclusion:
- Based on the test, there is enough evidence to reject the possibility that the vehicle-ownership distribution in the resort community is the same as that of the nation as a whole at the 0.05 level of significance.