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

orchestra answered 1 year ago

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