10.3.77 nonpooled t.test left-tailed

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.

\(\bar x_{1} = 18, s_1=3, n_1=20\)

\(\bar x_{2} = 21, s_2=5, n_2=15\)

(a).​ What are the correct hypotheses for a​ left-tailed test?

Since it is a left-tailed test, the correct hypothese is

\(H_0: \mu_1 =\mu_2\)

\(H_a: \mu_1 < \mu_2\)

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Find the test statistic. (Round to three decimal places as​ needed.)

First, we need to get the data from the question

x1 = 18
s1 = 3
n1 = 20
x2 = 21
s2 = 5
n2 = 15

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This is a nonpooled t-test, to compute the test statistic t we use the formular \(t= \frac{\bar x_1-\bar x_2}{\sqrt{s_1^2/n_1+s_2^2/n_2}}\)

t=(x1-x2)/sqrt((s1^2/n1)+(s2^2)/n2)
t

## [1] -2.06203

Round the answer to three decimal places

round(t,3)

## [1] -2.062

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Now determine the critical values. (Round to three decimal places as​ needed.)

We need to find degree of freedom by using the formula \(\Delta=\frac{[(s_1^2/n_1)+(s_2^2/n_2)]^2}{\frac{(s_1^2/n_1)^2}{n_1-1}+\frac{(s_2^2/n_2)^2}{n_2-1}}\)

deg = ((s1^2/n1) + (s2^2/n2))^2 / ( (s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1) )

Round down to nearest interger

deg = floor(deg)

Since this is a left-tailed test and \(\alpha = .05\), the area to the left of \(\alpha\) is equal to \(\alpha\)

alpha = .05
t_critical = abs(qt(alpha,deg))
t_critical

## [1] 1.720743

Since we use abs() command so it return the positive value of \(t_{\alpha}\) Round the answer to three decimal places

round(t_critical,3)

## [1] 1.721

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Since our test statistic t = -2.062 < our critical value \(t_{\alpha}\) = -1.721, out test statistic lies in the rejected region. So we have enough evidence to reject null hypothesis.

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(b) The 90% confidence interval is from… to…​(Round to three decimal places as​ needed.)

90% confidence interval means \(\alpha = .1\) so our critical value \(=\alpha/2=.05\) \(t_{\alpha/2} = t_{critical value}\)

To find confidence interval we use the formular \((\bar x_1 - \bar x_2) \pm t_{\alpha/2}.\sqrt{s_1^2/n_1+s_2^2/n_2}\)

(x1-x2) + t_critical * sqrt(s1^2/n1+s2^2/n2)

## [1] -0.496531

(x1-x2) - t_critical * sqrt(s1^2/n1+s2^2/n2)

## [1] -5.503469

Round to three decimal places

round((x1-x2) + t_critical * sqrt(s1^2/n1+s2^2/n2),3)

## [1] -0.497

round((x1-x2) - t_critical * sqrt(s1^2/n1+s2^2/n2),3)

## [1] -5.503

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We finish a lot of complicated work by using R

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Hope that helps!