Question

A psychologist is interested in whether hypnosis affects brain dominance. Twelve college students from the freshmen class are randomly sampled for an experiment. The experiment has two conditions which are given on different days. In condition 1, the students are hypnotized and then given a test which measures the relative dominance of the right and left hemispheres. The higher the score, the more dominant is the right hemisphere. In condition 2, the same students are given the test again, only this time they are not hypnotized but are in their normal state of consciousness. The following scores are obtained:
Student 1 2 3 4 5 6 7 8 9 10 11
Condition 1 26 15 22 19 23 37 16 17 14 16 21
Condition 2 12 13 18 17 21 22 11 22 10 14 15

Answer the following questions and show your work for each mathematical question:

a. What is the (non-directional) alternative hypothesis?
b. What is the null hypothesis?
c. What is the p value?

Answer 1

Answer:

$(a)\ H_a : d_1 - d_2 \ne 0$

$(b)\ H_0 : d_1 - d_2 = 0$

$(c)\ p\ value = 0.0216$

Step-by-step explanation:

Given

The above data

The question says 12 college students were sampled. However, data for 11 students were provided.

So, I will solve this question using 11 students i.e. n = 11

Solving (a): The alternative hypothesis

Let the relative dominance of the hypnotized students (condition 1) be $d_1$ and the relative dominance of the non-hypnotized students (condition 2) be $d_2$, respectively

The alternative hypothesis is the difference between these relative dominance not equal to 0.

So:

$H_a : d_1 - d_2 \ne 0$

Solving (b): The null hypothesis

The null hypothesis is the difference between these relative dominance equal to 0.

Using the same notation as (a), the null hypothesis is:

$H_0 = d_1 - d_2 = 0$

Solving (c): The p value

Using paired samples t test:

First, calculate the difference in scores

$i.e.\ Condition\ 1 - Condition\ 2$

The difference (d) is as follows:

$d: 14, 2, 4,2, 2, 15, 5, -5, 4, 2, 6$

Next, calculate the mean of the differences:

This is calculated as:

$\mu = \frac{\sum x}{n}$

$\mu = \frac{14+ 2+ 4+2+ 2+ 15+ 5 -5+ 4+ 2+ 6}{11}$

$\mu = \frac{51}{11}$

$\mu = 4.6364$

Next, calculate the standard difference of the differences.

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n-1}}$

So:

$\sigma = \sqrt{\frac{(14 - 4.6364)^2+ (2- 4.6364)^2+ (4- 4.6364)^2+ ............ (6- 4.6364)^2}{11-1}}$

$\sigma = \sqrt{\frac{318.54545456}{10}}$

$\sigma = \sqrt{31.854545456}$

$\sigma = 5.64398$  

 

Next, calculate the standard error of the mean difference.

This is calculated as:

$SE = \frac{\sigma}{\sqrt{n}}$

$SE = \frac{5.64398}{\sqrt{11}}$

$SE = \frac{5.64398}{3.3166}$

$SE = 1.70173$

Next, calculate the degree of freedom (df)

$df = n -1$

$df = 11 -1$

$df = 10$

Lastly, calculate the test statistics (t)

This is calculated as:

$t = \frac{\mu}{SE}$

$t = \frac{4.6364}{1.70173}$

$t = 2.72452$

$t \approx 2.72$

From the t distribution table:

Where $df = 10$ and $t = 2.72$

We have:

$P(t > 2.72) = 0.0108$

For a two-tailed test, the p value is

$p\ value = 2 * P(t > 2.72)$

$p\ value = 2 * 0.0108$

$p\ value = 0.0216$