Question

A researcher has developed a new drug designed to reduce blood pressure. In an experiment, 21 subjects were assigned randomly to the treatment group, and received the new experimental drug. The other 23 subjects were assigned to the control group, and received a standard, well known treatment. After a suitable period of time, the reduction in blood pressure for each subject was recorded. A summary of these data is:

                                                                                    n          x?         s

                        Treatment group (new drug):               21        23.48   8.01

                        Control group (old drug):                     23        18.52   7.15

A 99% confidence interval for the difference in reduction of blood pressure between these

two drugs is:

The researcher suspects that the new drug results in greater average reduction in blood

pressure than the old drug does. What would be the null and alternative hypothesis?

Based on this data, compute the two-sample t statistic.

The P-value for the test described is?

Show work please

Answer 1

We need to construct the 99% confidence interval for the difference between the population means μ1​−μ2​, for the case that the population standard deviations are not known.

The following information has been provided about each of the samples

Sample Mean 1 (Xˉ1​) =	23.48
Sample Standard Deviation 1 (s1​) =	8.01
Sample Size 1 (N_1) =	21
Sample Mean 2 (Xˉ2​) =	18.52
Sample Standard Deviation 2 (s_2) =	7.15
Sample Size 2 (N_2) =	23

Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are

df = n_1 + n_2 -2 = 21 + 23 - 2 = 42

The critical value for α=0.01 and df = 42 degrees of freedom is

The corresponding confidence interval is computed as shown below:

Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows:

Since we assume that the population variances are equal, the standard error is computed as follows:

Now, we finally compute the confidence interval:

Therefore, based on the data provided, the 99% confidence interval for the difference between the population means μ1​−μ2​ is −1.206<μ1​−μ2​<11.126,

which indicates that we are 99% confident that the true difference between population means is contained by the interval (-1.206, 11.126)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2

Ha: μ1​ < μ2​

(2) Rejection Region

Based on the information provided, the significance level is α=.01,

and the degrees of freedom are df = 42.

Hence, it is found that the critical value for this left-tailed test is t_c = -2.418

, for α=.01 and df = 42.

The rejection region for this left-tailed test is

R={t:t<−2.418}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that

t=2.17≥tc​=−2.418,

it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p = 0.9822,

and since p = 0.9822≥.01,

it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the .01 significance level.

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