Question

3. A study investigated survival rates for in-hospital patients who suffered cardiac arrest. Among 58,583 patients who has cardiac arrest during the day, 11,604 survived. Among 28,155 patients who suffered cardiac arrest during the night, 5350 survived. A physician claims that the survival rate of those who suffer cardiac arrest during the day is higher than the survival rate of those who suffer cardiac arrest during the night.

   1. a) Perform a complete hypothesis test to test their claim at the 5% level of significance. Use the P-value method.

   2. b) Construct a 90% confidence interval on the difference of proportions. Explain how your CI comes to the same conclusion as part a).

Answer 1

A physician claims that the survival rate of those who suffer cardiac arrest during the day is higher than the survival rate of those who suffer cardiac arrest during the night.

Here sample sizes are very large so we can use two sample proportion test :

Let's write the null hypothesis and the alternative hypothesis:

Let's used minitab:

Step 1: Click on Stat >>> Basic Statistics >>>2 Proportions...

Step 2: Select Summarized data

Fill the given information

Look the following picture ...

Then click on Option:

Look the following image:

Then click on OK again click on Ok

So we get the following output

From the above output

z = 2.80, p-value = 0.003

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.003 < 0.05 so we used first rule.

That is we reject the null hypothesis

Conclusion: At 5% level of significance there are sufficient evidence to say that the sample data indicates that the survival rate of those who suffer cardiac arrest during the day is higher than the survival rate of those who suffer cardiac arrest during the night.

b) Construct a 90% confidence interval on the difference of proportions. Explain how your CI comes to the same conclusion as part a).

In part b) after clicking option plug 90 instead of 95 and select not equal.

Look the following image

Then click on OK again click on Ok

So we get the following output

From the above output the 90% confidence interval on the difference of proportions is (0.00335458, 0.0127622)

since the lower limit of this confidence interval is greater than zero therefore we support the above claim.