10.5.149 paired t-test two-tailed
The null hypothesis is \(H_0:\mu_1=\mu_2\) and the alternative hypothesis is as specified. The provided data are from a simple random paired sample from the two populations under consideration. Use the paired t-test to perform the required hypothesis test at the 10% significance level.
\(H_a:\mu_1 \neq \mu_2\)
Find the test statistic. Use population 1 minus population 2 as the difference. (Round to three decimal places as needed.)
First, we need to get the data from the question
population1 <- c(5, 19, 20, 21, 7, 10, 9)
population2 <- c(3, 18, 17, 15, 3, 11, 6)Use population 1 minus population 2 as the difference.
We store difference between population 1 and population 2 in variable difference and run test statistic. Since this is two-tailed test and significant level = 10%, we have confidence level = .9 First approach use t.test()
difference = population1 - population2
t.test(difference, conf.level = .9)##
## One Sample t-test
##
## data: difference
## t = 3.0571, df = 6, p-value = 0.02231
## alternative hypothesis: true mean is not equal to 0
## 90 percent confidence interval:
## 0.9369806 4.2058765
## sample estimates:
## mean of x
## 2.571429Round to 3 decimal places
print(t.test(difference, conf.level = .9),6)##
## One Sample t-test
##
## data: difference
## t = 3.057, df = 6, p-value = 0.0223
## alternative hypothesis: true mean is not equal to 0
## 90 percent confidence interval:
## 0.936981 4.205877
## sample estimates:
## mean of x
## 2.57143
Since we have 7 pairs of data, our degree of freedom is 7- 1 = 6 and we have two-tailed test with \(\alpha = .1\), the area to the left of negative critical value = \(\alpha/2\) so \(t_{\alpha/2}\)
round(qt(.1/2, 6), 3)## [1] -1.943
Since the test statistic value t = 3.057 is larger than the positive critical value \(t_{\alpha/2} = 1.943\), we have enough evidence to reject hypothesis.
Second appraach to find test statistic, we can use formula \(t=\frac{\bar d}{\frac{s_d}{\sqrt{n}}}\)
n = 7
mean(difference)/(sd(difference)/sqrt(n))## [1] 3.057148Round to 3 decimal places
round(mean(difference)/(sd(difference)/sqrt(n)),3)## [1] 3.057Hope that helps!