Question

For all Hypothesis test problems, please do the following; following the steps as discussed in class:

  1.         Is the data discrete or continuous?

2.        List the assumptions

3.        State the null and alternate hypothesis.

4.        Write down the proper test statistic, Show all calculations, i.e. impute the numbers. Set up Decision rules. Show the critical value.    Reject or Fail to reject.  

5.        State your conclusions.

Repeat problem 6. 6. 17 using a chi-square test to test the hypothesis

6.6.17    An experiment was conducted to compare two methods for operating a group family medical practice. Four hundred patients were randomly assigned to two groups: one group received the conventional direct contact with physicians, while the other group made first contact with a nurse-practitioner and were then referred to a physician if a physician's services were deemed necessary. At the conclusion of the experiment, the quality of each person's medical care was rated as satisfactory by an impartial medical observer in consultation with the patient. The results of the experiment are shown in the table. Do the data present sufficient evidence to indicate a difference in the proportions of satisfactory ratings for the two methods of patient care? Test using a=0.05.

                                                                                             Results

First Contact                                  Satisfactory                     Unsatisfactory                      Totals

Conventional                                            146                                    54                              200

Nurse Practitioner                                    159                                    41                              200

Answer 1

Answer:

Null Hypothesis H0: There is no association between type of contact and results

Alternative Hypothesis H1: Type of contact and results are associated.

Under H0, the expected frequencies are calculated as:

	Results ( Obs Freq. ( O )			Results ( Exp Freq. ( E )
First Contact	Satisfactory	Unsatisfactory	Total	Satisfactory	Unsatisfactory	Total
Conventional	146	54	200	152.5	47.5	200
Nurse Practitioner	159	41	200	152.5	47.5	200
Total	305	95	400	305	95	400
Calculation For Chi Square
	Obs freq (O)	Exp. Freq (E )	(O-E)^2/E
	146	152.5	0.277
	54	47.5	0.889
	159	152.5	0.277
	41	47.5	0.889
Total	400	400	2.333

Under H0, the test statistic is

(0-Ег-2333

Degrees of freedom = (r-1)(c-1) = (2-1)*(2-1) = 1

For 1 df, at 5% significance level, the critical value of chi square is 3.841

The P-Value is 0.1267

Since p value is greater than significance level. Fail to Reject H0.

Conclusion : Hence the type of contact and results are associated.

At 5% significance level for two sided test,the critical value of Z is 1.96

Therefore (Z_critical)² = 1.96² = 3.8416 = (Critical)

The calculations for Z statistic

The proportion p1 for staisfied by conventional patients p1 = 146/200 =0.73

The proprtion p2 for statisfied by nurse practioner p2 = 159/200 = 0.795

Under H0, the test statistic is

2 2

146-159-305 = 0.7625 X1+ X2 0.7625 nln2 200+200 200

-0.065 = 0.0426 0.73 0.795 ー-1.5258 V/0.7325 * 0.2375 (2, + 2页) 200 200

(Zobs)2 = (-1.5258)2-233

(Zobs )2 = (-1.5258)2ー2.33-)-(obs)

for the chi-square value, p = 0.1269

Fail to reject H0