Question

A hospital reported that the normal death rate for patients with extensive burns (more than 40% of skin area) has been significantly reduced by the use of new fluid plasma compresses. Before the new treatment, the mortality rate for extensive burn patients was about 60%. Using the new compresses, the hospital found that only 43 of 92 patients with extensive burns died. Use a 1% level of significance to test the claim that the mortality rate has dropped.

What is the level of significance?

State the null and alternate hypotheses.

What sampling distribution will you use? What assumptions are you making?

What is the value of the sample test statistic? (Round your answer to two decimal places.)

Find (or estimate) the P-value.

Sketch the sampling distribution and show the area corresponding to the P-value.

Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

Interpret your conclusion in the context of the application.

Answer 1

a ) Level of significance = = 1% = 0.01

b ) Hypothesis :

Null hypothesis :

Alternative hypothesis :

Left tailed test.

c ) The standard normal, since np > 5 and nq > 5

d) Test statistic -

Where , is sample proportion = 43/92 = 0.46739

P is hypothesized sample proportion = 0.6

n is sample size = 92

Test statistic = z = -2.60

e) P-value.

P-value for this left tailed test is ,

P-value = P( z < test statistic ) = P( z < -2.60 )

Using Excel function ,   =NORMSDIST( z )

P( z < -2.60 ) = NORMSDIST(-2.60 ) = 0.0047

P-value = 0.0047

f)

Decision about null hypothesis :

Rule : Reject null hypothesis if P-value less than Level of significance

It is observed that, P-value ( 0.0047 ) is less than = 0.01

So reject null hypothesis..

Conclusion :

There is sufficient evidence to support the hospital report that death rate for patients with extensive burns (more than 40% of skin area) has been significantly reduced by the use of new fluid plasma compresses