Researchers are comparing the average daily screen time (in minutes) for individuals in three different income groups. Independent random samples of $6$ individuals were taken from each group. The calculated test statistic is $F=18.56$ for testing whether the mean daily screen time differs across the income groups. Assume normality of the populations and equal population variances. With a significance level of $\alpha =0.05$, interpret the results of the hypothesis test.

Two independent samples are drawn to compare variability.

Sample A: {$10$, $12$, $9$, $11$, $13$, $10$, $14$, $8$, $12$}

Sample B: {$7$, $9$, $6$, $8$, $7$, $9$, $5$, $8$}

Use the "count five" test: compute medians, absolute deviations from medians, maxima of absolute deviations $M_1$, $M_2$, counts $c_1$, $c_2$ where $c_1=\#\{d_{1i}:d_{1i}>M_2\}$ and $c_2=\#\{d_{2j}:d_{2j}>M_1\}$, and reject the null hypothesis (equal variances) if $\max (c_1,c_2)\ge 5$. Which conclusion is correct?

Two manufacturing plants produce the same component. A random sample from plant X yields $n_1=25$, ${\bar{x}}_1=45.2$, $s_1=5.4$. A random sample from plant Y yields $n_2=30$, ${\bar{x}}_2=42.6$, $s_2=6.1$. Assume the populations are approximately normal with equal variances. Construct a $95\%$ confidence interval for ${\mu }_1-{\mu }_2$.

A researcher is comparing the average recovery times for two medications. Test the claim that the mean recovery time for medication A is less than that for medication B at the $\alpha =0.05$ significance level. Assume independent random samples and equal population variances. Sample statistics: ${\bar{x}}_A=4.2$ days, $s_A=1.1$, $n_A=15$; ${\bar{x}}_B=5.0$ days, $s_B=0.9$, $n_B=12$.

Find the critical value for a left-tailed two-sample $t$-test with equal variances, given the following: , $\alpha =0.05$, $n_1=25$, $n_2=28$. Assume random, independent samples and normal populations.