Question
In: Statistics and Probability
In May 2004, a Gallup Poll of adults’ attitudes toward Health Maintenance Organization (HMOs) found that...
In May 2004, a Gallup Poll of adults’ attitudes toward Health Maintenance Organization (HMOs) found that 41% of adults had little or no confidence in HMOs, 38% had some confidence, 17% had a great deal or quite a lot of confidence, and 4% had no opinion (USA TODAY, June 22, 2004). Let us denote these outcomes as L, S, G, and N, respectively. A recent random sample of 500 adults yielded the frequency distribution given in the following table. Response L S G N Frequency 212 198 82 8
a. Determine the rejection region.
b. Using the 2.5% significance level, can you conclude that the current distribution of opinions differs from the distribution of May 2004?
Solutions
Expert Solution
| Observed | Expected | Difference | Difference Sq. | Diff. Sq. / Exp Fr. | |
| L | 212 | 205 | 7.00 | 49.00 | 0.24 |
| S | 198 | 190 | 8.00 | 64.00 | 0.34 |
| G | 82 | 85 | -3.00 | 9.00 | 0.11 |
| N | 8 | 20 | -12.00 | 144.00 | 7.20 |
Chi squared equals 7.882 with 3 degrees of freedom.
CRITICAL VALUE=0.1848
If your chi-square calculated value is greater than the chi-square critical value, then you reject your null hypothesis. If your chi-square calculated value is less than the chi-square critical value, then you "fail to reject" your null hypothesis.
Null hypothesis: Assumes that there is no association between the two variables.
Alternative hypothesis: Assumes that there is an association between the two variables
- REJECTION REGION IS GREATER THAN 0.1848
- CALCULATED>CRITICAL, REJECT NULL HYPOTHESIS SO there is an association between the two OBSERVED AND EXPECTED
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