Question

In: Statistics and Probability

A hospital claims that 60% of admissions related to cardiac arrest are discharged home alive. A...

A hospital claims that 60% of admissions related to cardiac arrest are discharged home alive. A researcher is suspicious of the claim and thinks that the proportion of survival to hospital discharge following cardiac arrest is lower than 60%. A random sample of 120 cardiac arrest cases show that only 60 survived. Is there sufficient evidence to show that the true population proportion is less than 60%? The test statistic is z=-2.24. What do you conclude?

A.Nothing, the conditions to conduct a test were not met.

B. There is not adequate evidence to show that the population proportion is different than 60%

C.There is adequate evidence to show that the population proportion is greater than 60%

D.There is adequate evidence to show that the population proportion is less than 60%

Solutions

Expert Solution

Solution:

From the given information ,

the hypothesis can be written as ,

H0 : p = 60%

H1 : p < 60%

n = 120

x = 60

The test statistic calculated is

z = -2.24

< sign in H1 indicates that the test is left tailed.

p value = P(Z < z)

= P(Z < -2.24)

= 0.0125

Now , we know that if the p value is less than the significance level , then we reject the null hypothesis H0 and the test is said to be significant. otherwise we fail to reject the null hypothesis and test is said to be not significant.

In the given example , the significance level is not given.

In this case , we generally take 0.05 significance level.

p value = 0.0125 < = 0.05

We reject the null hypothesis H0 and conclude that the population proportion is less than 60%

Answer:

D.There is adequate evidence to show that the population proportion is less than 60%


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