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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