Question

A research firm reported that 14% of those surveyed described their health as poor, 30% as good, 36% as very good, and 20% as excellent. A health professional in Chicago wanted to determine if people in Chicago had similar feelings toward their health and collected the data below. Use hypothesis testing to determine if these data agree with the survey report at α = 0.05.

Health status	Poor	Good	Very good	Excellent	Total
Count	70	175	225	130	600

What is the critical value?

a. 7.815

b. 11.143

c. 9.488

d. 9.348

Answer 1

Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147.

The entire test:

Chi-Square Goodness of Fit test

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.14, p2 =0.3, p3​=0.36, p4​=0.2 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Degrees of Freedom The number of degrees of freedom is df=n-1=4-1=3 (3) Test Statistics The Chi-Squared statistic is computed as follows: (4)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147. Then the rejection region for this test is R={χ2:χ2>7.8147}. (5)P-value The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 7.8147. The p-value is p=Pr(χ2>3.6806)=0.2981 (6) The decision about the null hypothesis Since it is observed that χ2=3.6806≤χc2​=7.8147, it is then concluded that the null hypothesis is not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.

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