Question

an educational psychologist is studying two methods of tutoring young children: method a and method b. she wants to test whether there is any difference in puzzle - solving skills between children tutored by one method versus children tutored by the other. independent samples of 13 children who were tutored using method a and 9 children who were tutored using method b were chosen at random. the method a children took a mean of 35 minutes to solve a certain puzzle with a standard deviation of 3 minutes. the method b children took a mean of 42 minutes to solve the same puzzle with a standard deviation of 7 minutes. assume that the two populations of completion times are normally distributed and that the population variances are equal. construct a 99% confidence interval for the difference μ₁ - μ₂ between the mean puzzle - solving times for method a children (μ₁) and for method b children (μ₂). then find the lower limit and upper limit of the 99% confidence interval. carry your intermediate computations to at least three decimal places. round your responses to at least two decimal places. (if necessary, consult a list of formulas.) lower limit: upper limit:

Explanation:

Step1: Calculate pooled variance

The formula for pooled - variance $s_p^2$ is $s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}$, where $n_1 = 13$, $s_1 = 3$, $n_2 = 9$, $s_2 = 7$.
\[

$$\begin{align*} s_p^2&=\frac{(13 - 1)\times3^2+(9 - 1)\times7^2}{13 + 9-2}\\ &=\frac{12\times9 + 8\times49}{20}\\ &=\frac{108+392}{20}\\ &=\frac{500}{20}\\ & = 25 \end{align*}$$

\]

Step2: Find the degrees of freedom and critical value

The degrees of freedom $df=n_1 + n_2-2=13 + 9-2 = 20$. For a 99% confidence interval, the significance level $\alpha=0.01$, and $\alpha/2=0.005$. Looking up in the t - distribution table, the critical value $t_{\alpha/2,df}=t_{0.005,20}=2.845$.

Step3: Calculate the confidence interval

The formula for the confidence interval for $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)\pm t_{\alpha/2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$, where $\bar{x}_1 = 35$, $\bar{x}_2 = 42$, $s_p=\sqrt{25}=5$, $n_1 = 13$, $n_2 = 9$.
First, calculate $\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{\frac{1}{13}+\frac{1}{9}}=\sqrt{\frac{9 + 13}{117}}=\sqrt{\frac{22}{117}}\approx0.434$.
Then, $t_{\alpha/2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=2.845\times5\times0.434 = 2.845\times2.17\approx6.174$.
The lower limit is $(\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=(35 - 42)-6.174=-7 - 6.174=-13.174\approx - 13.17$.
The upper limit is $(\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=(35 - 42)+6.174=-7 + 6.174=-0.826\approx - 0.83$.

Answer:

Lower limit: $-13.17$
Upper limit: $-0.83$