Question

The Australian Medical Association believed that the Health Minister's recent statement claiming that 75% of doctors supported the reforms to Medicare was incorrect. It thought that the actual support for the reforms was lower than this. The Association's President suggested the best way to test this was to survey 150 members, selected through a random sample, on the issue. She indicated that the Association would be prepared to accept a Type I error probability of 0.05.

1. State the direction of the alternative hypothesis for the test. Type gt (greater than), ge (greater than or equal to), lt (less than), le (less than or equal to) or ne (not equal to) as appropriate in the box.

2. State, in absolute terms, the critical value as found in the tables in the textbook.

3. Determine the lower boundary of the region of non-rejection in terms of the sample proportion of respondents (as a % to two decimal places) in favour of the reforms. If there is no (theoretical) lower boundary, type lt in the box.

4. Determine the upper boundary of the region of non-rejection in terms of the sample proportion of respondents (as a % to two decimal places) in favour of the reforms. If there is no (theoretical) upper boundary, type gt in the box.

5. If 102 of the survey participants indicated support for the reforms, is the null hypothesis rejected for this test? Type yes or no.

6. Disregarding your answer for 5, if the null hypothesis was rejected, could the Association claim that the Health Minister's assertion is incorrect at the 5% level of significance?

Answer 1

Solution:

1. the direction of the alternative hypothesis for the test. greater than, greater than or equal to, less than, less than or equal to or not equal to are:

The Australian Medical Association claim is that 75% of doctors supported the reforms to Medicare is incorrect statement.

So, the direction of the alternative hypothesis for the test is not equal to.

2. in absolute terms, the critical value as found is:

For Type I error probability of 0.05, and two tail test, the critical values of z are found at

0.05/2 = 0.025 and

(1 - 0.05/2) = 0.975

The critical values of z are -1.96 and 1.96

3. the lower boundary of the region of non-rejection in terms of the sample proportion of respondents :

Hypothesized proportion, p = 0.75

Standard error of Proportion = sqrt [p(1-p)/n]

= sqrt [0.75(1-0.75)/150]

= 0.03535

Lower boundary of the region of non-rejection = 0.75 - 1.96 * 0.03535 = 0.680

4. the upper boundary of the region of non-rejection in terms of the sample proportion of respondents ​​​​​​​:

Upper boundary of the region of non-rejection = 0.75 + 1.96 * 0.03535 = 0.819

5.If 102 of the survey participants indicated support for the reforms, is the null hypothesis rejected for this test​​​​​​​ :

Sample proportion = 102 / 150 = 0.68

As, 0.68 lies in the non-rejection region (0.631, 0.769), the null hypothesis is not rejected.

6. if the null hypothesis was rejected, could the Association claim that the Health Minister's assertion is incorrect at the 5% level of significance:

Yes,  

if the null hypothesis was rejected, proportion of doctors supported the reforms to Medicare is not 75% and the Association can claim that the Health Minister's assertion is incorrect at the 5% level of significance