Question

4.12 To determine whether training in a series of workshops on creative thinking ncreases 1Q 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 10(%) for the treatment group equals 110, and the sample mean IQ (X2) 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 stan- dardized effect size, construct a 99 percent confidence interval for the true popula- tion mean difference, and interpret these estimates.

Answer 1

Solution : ( a )

$\mathrm{H_{0}}:\mu _{1}-\mu _{2}\leq 0$

$\mathrm{H_{1}}:\mu _{1}-\mu _{2}> 0$

$\mathrm{Reject \:H_{0}\:at\:the\:0.01\:level\:of\:significance\:if\:t\geq 2.390,\:given\:df=35+35-2=70-2=68}$

$\mathrm{t=\left ( \frac{\bar{X}-\mu_{\mathrm{hyp}} }{s_{\bar{X}}} \right )}=\frac{110-108}{1.80}=\frac{2}{1.8}=1.11$

$\mathrm{Retain \:H_{0}\:at\:the\:0.01\:significance\:because\:t=1.11\:is\:less\:than\: 2.390}$

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Solution : ( b )

$\mathrm{It \:is\:not\:appropriate\:since\:the\:null\:hypothesis\:was\:retained.}$