Question 1206153: Nationally, 60% of American own American cars, 29% of Americans own Asian cars, and the remaing Americans own non-Asian foreign cars. A survey of 206 people in California finds that 51% own American cars, 31% own Asian cars, and the remaining own non-Asian foreign cars. When testing (at the 10% level of significance) whether the proportions in California are the same as the national proportions, what is the test statistic (please round your answer to 3 decimal places, show work)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! ## Setting Up the Chi-Squared Test
**Understanding the Problem:**
We want to test if the distribution of car ownership in California differs from the national distribution. This is a classic chi-squared test for goodness of fit.
**Null Hypothesis (H₀):** The distribution of car ownership in California is the same as the national distribution.
**Alternative Hypothesis (H₁):** The distribution of car ownership in California is different from the national distribution.
**Setting up the Contingency Table:**
| Car Type | National Proportion | California Proportion | Observed Frequency | Expected Frequency |
|---|---|---|---|---|
| American | 0.60 | 0.51 | 105 | 123.96 |
| Asian | 0.29 | 0.31 | 64 | 59.74 |
| Non-Asian Foreign | 0.11 | 0.18 | 37 | 22.29 |
| Total | 1.00 | 1.00 | 206 | 206 |
**Calculating Expected Frequencies:**
To calculate the expected frequencies, we multiply the total number of observations (206) by the national proportion for each car type. For example, the expected frequency for American cars in California is 206 * 0.60 = 123.96.
**Calculating the Chi-Squared Test Statistic:**
The chi-squared test statistic is calculated as:
```
χ² = Σ [(Observed Frequency - Expected Frequency)² / Expected Frequency]
```
Calculating this for each cell and summing them up, we get:
```
χ² = [(105 - 123.96)²/123.96] + [(64 - 59.74)²/59.74] + [(37 - 22.29)²/22.29]
```
**Calculating the Test Statistic:**
Using a calculator or statistical software, we find the chi-squared test statistic to be approximately **10.878**.
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