Question

14.12 To determine whether training in a series of workshops on creative thinking increases IQ scores, a total of 70 students are randomly divided into treatment and control groups of 35 each. After two months of training, the sample mean IQ 1X for the treatment group equals 110, and the sample mean IQ 2X for the control group equals 108. The estimated standard error equals 1.80. (a) Using t, test the null hypothesis at the .01 level of significance (b) If appropriate (because the null hypothesis has been rejected), estimate the standardized effect size, construct a 99 percent confidence interval for the true population mean difference, and interpret these estimates.

Answer 1

Answer:

reject H₀ , {0.8262, 3.1738} at 99% C I

Step-by-step explanation:

$H_0: \mu_1 = \mu_2\\H_1: \mu _1 > \mu _2$

critical value, $\alpha = 0.01$

degree of freedom: 35 - 1 = 34

critical value: 2.441

from the expression

$t = \frac{(x _1-x_2) - 0}{\sqrt{\frac{\sigma^{2}_x }{n_1} +\frac{\sigma^{2}_x }{n_2} } }$

$t = \frac{(110-108) - 0}{\sqrt{\frac{1.8^{2} }{35} +\frac{1.8^{2}}{35} } }$

t = 4.648 > 2.441

H₀ is rejected because t > critical value

(b)

in the second scenario

$I: (x_1 - x_2) +/- t\sqrt{\frac{\sigma^{2}_x }{n_1} +\frac{\sigma^{2}_x }{n_2}$

1 - 0.99 = 0.01

$\frac{\alpha }{2} = \frac{0.01}{2}= 0.005$

degree of freedom = 34

t = 2.728

substituting in the above formula

we have

$I: (110 - 108) +/- 2.728\sqrt{\frac{1.80^{2}}{35} +\frac{1.8^{2}}{35}}$

$I: 2 +/- 1.1738$

$I:[0.8262, 3.1738]$

we can see the difference of the means is between 0.8262 and 3.1738 at 99% confidence