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
Total 100
It would indeed depend upon the "sample" included in the survey. Resort area? Those people would be wealthier by far than in the "barrio" for example. If at the end of this post you tried to "copy and paste" it does not usually work here. You will have to type the rest out.
Sra
In a test of the independence of two variables, one of the variables has two possible categories and the other has three possible categories. What will be the critical value of chi-square if the test is to be carried out at the o.025 level? At the 0.05 level?
To determine if we can reject the possibility that the vehicle-ownership distribution in the resort community differs from that of the nation as a whole, we can use a hypothesis test. Specifically, we will use a chi-square test for independence.
First, let's state our null and alternative hypotheses:
Null Hypothesis (H0): The vehicle-ownership distribution in the resort community is the same as the national distribution.
Alternative Hypothesis (Ha): The vehicle-ownership distribution in the resort community differs from the national distribution.
Next, we need to calculate the expected frequencies for each category based on the national distribution. Since we are given the percentages of households in the nation, we can use those percentages to calculate the expected frequencies.
The expected frequencies can be found by multiplying the total number of households (100) by the respective percentages for each category:
Expected Frequency for 0 vehicles = 100 * 0.103 = 10.3
Expected Frequency for 1 vehicle = 100 * (1 - 0.342 - 0.171) = 44.7
Expected Frequency for 2 vehicles = 100 * 0.342 = 34.2
Expected Frequency for 3 or more vehicles = 100 * 0.171 = 17.1
Now, let's set up a contingency table with observed and expected frequencies:
Number of Vehicles Owned | Observed Frequency | Expected Frequency
-------------------------------------------------------------
0 | 20 | 10.3
1 | 35 | 44.7
2 | 23 | 34.2
3 or more | 22 | 17.1
We can now calculate the chi-square test statistic using the formula:
χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]
Calculation:
χ² = [(20 - 10.3)² / 10.3] + [(35 - 44.7)² / 44.7] + [(23 - 34.2)² / 34.2] + [(22 - 17.1)² / 17.1]
Once we calculate the chi-square test statistic, we can compare it to the critical value from the chi-square distribution table at the given level of significance (0.05) and degrees of freedom (df = number of categories - 1).
If the calculated chi-square value is greater than the critical value, we can reject the null hypothesis and conclude that the vehicle-ownership distribution in the resort community differs from the national distribution. Otherwise, if the calculated chi-square value is less than or equal to the critical value, we fail to reject the null hypothesis.
I will now perform the calculations to find the chi-square test statistic and check if we can reject the null hypothesis.