Question

Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled​ t-test and the nonpooled​ t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.
X1 = 11, S1 = 5, n1 = 25, x2 = 10, S2 = 4, n2 = 20
A) What are the correct hypotheses for a​right-tailed test? α = 0.05
i. Compute the test statistic.
ii. Determine the​ P-value.
B) The 90​% confidence interval is from _______ to ________

Answer 1

Using the t-distribution, we have that:

a)

• The null hypothesis is $H_0: x_1 - x_2 \leq 0$.

• The alternative hypothesis is $H_1: x_1 - x_2 > 0$

i) The test statistic is t = 0.746.

ii) The p-value is of 0.2299.

b) The 90​% confidence interval is from -1.25 to 3.25.

Item a:

For a right-tailed test, we test if $x_1$ is greater than $x_2$, hence:

• The null hypothesis is $H_0: x_1 - x_2 \leq 0$.

• The alternative hypothesis is $H_1: x_1 - x_2 > 0$

The standard errors are:

$S_{e1} = \frac{5}{\sqrt{25}} = 1$

$S_{e2} = \frac{4}{\sqrt{20}} = 0.8944$

The distribution of the difference has mean and standard error given by:

$\overline{x} = x_1 - x_2 = 11 - 10 = 1$

$s = \sqrt{S_{e1}^2 + S_{e2}^2} = \sqrt{1^2 + 0.8944^2} = 1.34$

We have the standard deviation for the samples, hence, the t-distribution is used.

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{1 - 0}{1.34}$

$t = 0.746$

The test statistic is t = 0.746.

The p-value is found using a t-distribution calculator, with t = 0.746, 25 + 20 - 2 = 43 df and 0.05 significance level.

• Using the calculator, it is of 0.2299.

Item b:

The critical value for a 90% confidence interval with 43 df is $t = 1.6811$.

The interval is:

$\overline{x} \pm ts$

Hence:

$\overline{x} - ts = 1 - 1.6811(1.34) = -1.25$

$\overline{x} + ts = 1 + 1.6811(1.34) = 3.25$

The 90​% confidence interval is from -1.25 to 3.25.

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