Are these independent samples? Yes, since the samples from the two machines are not related. Are these large samples or a normal population? We have \(n_1\lt 30\) and \(n_2\lt 30\). We do not have large enough samples, and thus we need to check the normality assumption from both populations. Let's take a look at the normality plots for this data: Normal Probability Plot for New Machine  Normal Probability Plot for Old Machine From the normal probability plots, we conclude that both populations may come from normal distributions. Remember the plots do not indicate that they DO come from a normal distribution. It only shows if there are clear violations. We should proceed with caution. Do the populations have equal variance? No information allows us to assume they are equal. We can use our rule of thumb to see if they are “close.” They are not that different as \(\dfrac{s_1}{s_2}=\dfrac{0.683}{0.750}=0.91\) is quite close to 1. This assumption does not seem to be violated. We can thus proceed with the pooled t-test. Let \(\mu_1\) denote the mean for the new machine and \(\mu_2\) denote the mean for the old machine. The null hypothesis is that there is no difference in the two population means, i.e. \(H_0\colon \mu_1-\mu_2=0\) The alternative is that the new machine is faster, i.e. \(H_a\colon \mu_1-\mu_2<0\) The significance level is 5%. Since we may assume the population variances are equal, we first have to calculate the pooled standard deviation: \begin{align} s_p&=\sqrt{\frac{(n_1-1)s^2_1+(n_2-1)s^2_2}{n_1+n_2-2}}\\ &=\sqrt{\frac{(10-1)(0.683)^2+(10-1)(0.750)^2}{10+10-2}}\\ &=\sqrt{\dfrac{9.261}{18}}\\ &=0.7173 \end{align} The test statistic is: \begin{align} t^*&=\dfrac{\bar{x}_1-\bar{x}_2-0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ &=\dfrac{42.14-43.23}{0.7173\sqrt{\frac{1}{10}+\frac{1}{10}}}\\&=-3.398 \end{align} The alternative is left-tailed so the critical value is the value \(a\) such that \(P(T<a)=0.05\), with \(10+10-2=18\) degrees of freedom. The critical value is -1.7341. The rejection region is \(t^*<-1.7341\). Our test statistic, -3.3978, is in our rejection region, therefore, we reject the null hypothesis. With a significance level of 5%, we reject the null hypothesis and conclude there is enough evidence to suggest that the new machine is faster than the old machine. For the independent-measures t test, which of the following describes the pooled variance (whose symbol is )? The difference between the standard deviations of the two samples A weighted average of the two sample variances (weighted by the sample sizes) An estimate of the standard distance between the difference in sample means (M₁ – M₂) and the difference in the corresponding population means (μ₁ – μ₂) The variance across all the data values when both samples are pooled together For the independent-measures t test, which of the following describes the estimated standard error of the difference in sample means (whose symbol is )? An estimate of the standard distance between the difference in sample means (M₁ – M₂) and the difference in the corresponding population means (μ₁ – μ₂) The difference between the standard deviations of the two samples The variance across all the data values when both samples are pooled together A weighted average of the two sample variances (weighted by the sample sizes) In calculating , you typically first need to calculate . is the value used in the denominator of the t statistic for the independent-measures t test. Suppose you conduct a study using an independent-measures research design, and you intend to use the independent-measures t test to test whether the means of the two independent populations are the same. The following is a table of the information you gather. Fill in any missing values.
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Differentiating pooled variance and the estimated standard error of the difference in sample means
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