How to find the Z and P values for these percentages?

Question

This is the question exactly from my homework:

In a study of the effectiveness of a fabric device that acts like a support stocking for a weak or damaged heart, 110 people who consented to treatment were assigned at random to either a standard treatment consisting of drugs or the experimental treatment that consisted of drugs plus surgery to install the stocking. After two years, 45% of the 60 patients receiving the stocking had improved and 36% of the patients receiving the standard treatment had improved. (Use a statistical computer package to calculate the P-value. Use p(experimental) − p(standard). Round your test statistic to two decimal places and your P-value to four decimal places.)

Please help! I don't understand how to find the z-score or the p-score of these percentages!

Answer 1

There are several possible methods for testing the null hypothesis that the population proportions of successes with the stocking and standard methods are equal. You have 27 successes out 60 for the stocking method and 18 out of 50 successes for the standard method.

You aren't specific about the tests shown in your textbook or covered in lectures for your course, so I can't discuss specifics. But I can show you output from two computer programs, since you were asked to use software.

Normal test from Minitab. Here is output from Minitab statistical software, which uses a z-statistic based on the two sample proportions of successes. The test statistic has a standard normal distribution under the null hypothesis that the methods do not differ. The result is $Z = 0.96$ so the P-value for a two-sided test is $P(|Z| > .96) = 2P(Z > .96) \approx 0.45 > .05.$ There are several normal tests that might apply, and this is one of them. ['Fisher's exact test', shown at the end, is a separate statistical method; you can google it if it is not in your textbook.]

Test and CI for Two Proportions
Method
p₁: proportion where Sample 1 = Event
p₂: proportion where Sample 2 = Event
Difference: p₁ - p₂
Descriptive Statistics
  Sample     N  Event   Sample p
Sample 1    60     27   0.450000
Sample 2    50     18   0.360000

Estimation for Difference
Difference  95% CI for
Difference
0.09    (-0.093160, 0.273160)
CI based on normal approximation

Test
Null hypothesis H₀: p₁ - p₂ = 0
Alternative hypothesis  H₁: p₁ - p₂ ≠ 0
              Method    Z-Value P-Value
Normal approximation    0.96    0.336
Fisher's exact                  0.436

Chi-squared test from R. Here is output from R statistical software, which uses a chi-squared test of a two-way table of succeses and failures. This test statistic $Q$ has a chi-squared distribution with 1 degree of freedom. Because $Q = 0.579$ rejects the null hypothesis for large values of $Q,$ the P-value is $P(Q > 0.579) \approx 0.045 > .05.% [There is a difference of opinion among statisticians whether the 'continuity correction' is appropriate here, but using it does not change the decision not to reject.]

> succ = c(27, 18);  subj=c(60,50)
> prop.test(succ,subj)

        2-sample test for equality of proportions with continuity correction

data:  succ out of subj
X-squared = 0.57946, df = 1, p-value = 0.4465
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.111493  0.291493
sample estimates:
prop 1 prop 2 
  0.45   0.36 

As you can see, neither method finds a significant difference between the methods of treatment at the 5% level (both P-values above 5%). I hope you can match the output from one of these two statistical programs with material covered in your course.