Question

Suppose we are interested in testing null hypothesis that difference of two population means is the same. Consider two samples collected from a normal population.

x	4.2	4.5	4.9	5.6	5.7	5.9	6.1	6.3	6.8	7.1	7.3	7.8	8.4	8.6	9.1	9.7	10.2
y	5.6	5.9	6.5	7.8	8.5	9.3

We need to testH0:x=yvsHa:x̸=yWe can use R-command t.test to compute the p-value and the confidence interval.> x = c ( 4.2 ,4.5,4.9,5.6,5.7 ,5.9, 6.1, 6.3 , 6.8 , 7.1 , 7.3 ,7.8,8.4,8.6 , 9.1, 9.7, 10.2)#entering data > y = c ( 5.6 ,5.9,6.5,7.8 ,8.5, 9.3)#entering data> x[1] 4.2 4.5 4.9 5.6 5.7 5.9 6.1 6.3 6.8 7.1 7.3 7.8 8.4 8.6 9.1 9.7 10.2> y[1] 5.6 5.9 6.5 7.8 8.5 9.3> t.test(x,y,mu=0,var.equal=F,conf.level=0.95)# Ho:mux

Answer 1

R Output

> x = c ( 4.2 ,4.5,4.9,5.6,5.7 ,5.9, 6.1, 6.3 , 6.8 , 7.1 , 7.3 ,7.8,8.4,8.6 , 9.1, 9.7, 10.2)
> y = c ( 5.6 ,5.9,6.5,7.8 ,8.5, 9.3)
> t.test(x,y,mu=0,var.equal=F,conf.level=0.95)

   Welch Two Sample t-test

data: x and y
t = -0.41807, df = 10.565, p-value = 0.6843
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-1.973709 1.346258
sample estimates:
mean of x mean of y
6.952941 7.266667

The following null and alternative hypotheses need to be tested:

Ho: μ_x​ =μ_y

Ha: μ_x ≠ μ_y​

This corresponds to a two-tailed test.

Since it is assumed that the population variances are unequal, the t-statistic is:

t = -0.41807

Using the P-value approach: The p-value is p-value = 0.6843, and since p-value = 0.6843 > 0.05, it is concluded that the null hypothesis is not rejected. and we conclude that μ_x​ =μ_y

95 percent confidence interval:
-1.973709 1.346258