Question

A study was done on 40 female patients following a new treatment for cardio-vascular disease (CVD).   Doctors measured the increase in exercise capacity (in minutes) over a 6-week period. The conventional treatment had produced an average increase of 2 minutes. Researchers wish to claim that the new treatment will increase the mean exercise capacity more than the conventional treatment. The data yielded x=2.3 and S =1.2

1. Conduct a 5-step test of hypotheses to test the researchers’ claim, use a = .05.
2. Find a 90% confidence interval for the population mean

Answer 1

a. Let µ be the true mean or true average of increase in exercise capacity.

The conventional treatment had produced an average increase of 2 minutes i.e  µ=2.

Researchers wish to claim that  mean exercise capacity more than the conventional treatment i.e  µ>2.

So we need to test the null hypothesis H0:  µ=2 vs the alternative hypothesis H1: µ>2.

The test is for one-sample mean where the population standard deviation is unknown.

The test statistic is:

t={√n*(X̄-µ0)}/s

where n is the sample size=40

X̄=sample mean=2.3

µ0=hypothesized mean=2

s=sample standard deviation =1.2

So, test statistic is = {√40*(2.3-2)}/1.2 =1.5811

Critical value and p-value:

The test is right-sided so, the critical value is t(alpha=0.05, degrees of freedom=n-1) i.e t(0.05,39) i.e t value for alpha=0.05 with degrees of freedom=39.

Now,Critical value= t(0.05,39)=1.685 [From t table]

The p-value for one-sided test for observed t=1.5811 with degrees of freedom=39 is P( t>1.5811)=0.0610

p-value=0.0610

Decision :

Decision rule:

If test statistic t is > critical t , then will reject the null hypothesis. Otherwise we will fail to reject the null hypothesis.

For p-value, if p-value is < alpha,then will reject the null hypothesis. Otherwise we will fail to reject the null hypothesis.

Reject/ Not reject:

Critical value approach

The test statistic =1.5811 is < critical value =1.685

Decision: Fail to reject the null hypothesis.

p-value approach

The p-value=0.0610 is > alpha=0.05.

Decision: Fail to reject the null hypothesis.

Hence we fail to reject the null hypothesis at alpha=0.05

Conclusion: We don't have enough evidence to conclude that the new treatment will increase the mean exercise capacity more than the conventional treatment. Hence claim is not true.

b. The confidence interval for population mean or true mean µ for unknown population standard deviation is given by:

X̄± t(alpha/2,n-1) * s/√n

i.e X̄ - t(alpha/2,n-1) * s/√n< µ <X̄ + t(alpha/2,n-1) * s/√n

where,

where n is the sample size=40

X̄=sample mean=2.3

s=sample standard deviation =1.2

t(alpha/2,n-1) is the two-sided t-value for degrees of freedom (n-1) with level of significance alpha

For 90% CI, alpha=1-0.90=0.1

alpha/2=0.1/2=0.05

So, here t(alpha/2,n-1)=t(0.05,39)=1.685 [ From part (a)]

Hence, 90% CI for µ is:

X̄± t(alpha/2,n-1) * s/√n

i.e 2.3 ± 1.685 *1.2/√40

i.e 2.3 ± 0.3197

i.e (1.9803, 2.6197)

So the 90% confidence interval for population mean µ is:

1.9803 < µ <2.6197 (Ans)