Chapter 10: Q.7 (page 663)

A “subliminal” message is below our threshold of awareness but may nonetheless influence us. Can subliminal messages help students learn math? A group of 18 students who had failed the mathematics part of the City University of New York Skills Assessment Test agreed to participate in a study to find out. All received a daily subliminal message, flashed on a screen too rapidly to be consciously read. The treatment group of 10 students (assigned at random) was exposed to “Each day I am getting better in math.” The control group of $8$ students was exposed to a neutral message, “People are walking on the street.” All $18$ students participated in a summer program designed to raise their math skills, and all took the assessment test again at the end of the program. The table below gives data on the subjects’ scores before and after the program.
(a) The Fathom boxplots below display the differences in pretest and posttest scores for the students in the control (Cdiff) and treatment (Tdiff) groups. Write a few sentences comparing the performance of these two groups.
(b) Construct and interpret a $90 %$ confidence interval for the mean difference in gains between treatment and control.
(c) Can we generalize these results to the population of all students who failed the mathematics part of the City University of New York Skills Assessment Test? Why or why not?

Short Answer

Expert verified

a). The distribution of the control group seems to be right-skewed, whereas the distribution of the treatment group seems symmetric.

b). There are $90 %$ confident that the mean difference is between $0.0306$ and $6.2694$ .

c). No

Step by step solution

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

The center of the treatment group is greater than the treatment group for the control group, the reason is that the median lies more to the right of the box plot.

The spread for both groups looks to be about the same because the width between the whiskers of the box plots is the approximately same for both box pots.

The distribution of the control group seems to be right-skewed, whereas the distribution of the treatment group seems symmetric.

Part (b) Step 1: Given Information

The mean is

${\bar{x}}_{1} = 11.4$

{\bar{x}}_{2} = 8.25

Part (b) Step 2: Explanation

Lower and upper limit is

$({\bar{x}}_{1} - {\bar{x}}_{2}) - t_{α / 2} \times \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$

$({\bar{x}}_{1} - {\bar{x}}_{2}) + t_{α / 2} \times \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$

The mean is the sum of all values divided by the number of values:

${\bar{x}}_{1} = 11.4$

${\bar{x}}_{2} = 8.25$

The variance is

$s_{1} = 3.1693$

$s_{2} = 3.6936$

$c = 90 %$

Part (b) Step 3: Explanation

The degrees of freedom:

$d f = 7$

Determine $t_{c}$ with $d f = 7$ using table B:

$t_{c} = 1.895$

The endpoints of the confidence interval for $μ_{1} - μ_{2}$ are:

$({\bar{x}}_{1} - {\bar{x}}_{2}) - t_{α / 2} \times \sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$

$= (11.4 - 8.25) - 1.895 \times \sqrt{\frac{3 . 1693^{2}}{10} + \frac{3 . 6936^{2}}{8}}$

$= 0.0306$

$({\bar{x}}_{1} - {\bar{x}}_{2}) + t_{α / 2} \times \sqrt{\frac{s_{1}^{1}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}$ $= (11.4 - 8.25) + 1.895 \times \sqrt{\frac{3 . 1693^{2}}{10} + \frac{3 . 6936^{2}}{8}}$

=6.2694

Part (c) Step 1: Given Information

Part (c) Step 2: Explanation

No, because the students admit to participate in the study and therefore the sample is not a arbitrary sample. Although the arbitrary requirement is not satisfied, it is not possible to generalize to the all the population.