Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have [tex]n_{1} = 25[/tex], [tex]\bar{x}_{1} = 2.04[/tex], [tex]\sigma_{1} = 0.010[/tex] and [tex]n_{2} = 20[/tex], [tex]\bar{x}_{2} = 2.07[/tex], [tex]\sigma_{2} = 0.015[/tex]. The pooled estimate is given by
[tex]\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552[/tex]
a. We want to test [tex]H_{0}: \mu_{1}-\mu_{2} = 0[/tex] vs [tex]H_{1}: \mu_{1}-\mu_{2} \neq 0[/tex] (two-tailed alternative).
The test statistic is [tex]T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}[/tex] and the observed value is [tex]t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257[/tex]. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value [tex]t_{0}[/tex] falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by [tex]2P(T<-8.0257)\approx[/tex]0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
[tex](\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}[/tex], i.e.,
[tex]-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}[/tex]
where [tex]t_{0.025}[/tex] is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
[tex]-0.03\pm(2.0167)(0.012459)(0.3)[/tex], i.e.,
(-0.0225, -0.0375)
