A primatologist believes that rhesus monkeys possess curiosity. She reasons that, if this is true, then they should prefer novel stimulation to repetitive stimulation. An experiment is conducted in which 12 rhesus monkeys are randomly selected from the university colony and taught to press two bars. Pressing bar 1 always produces the same sound, whereas bar 2 produces a novel sound each time it is pressed. After learning to press the bars, the monkeys are tested for 15 minutes, during which they have free access to both bars. The number of presses on each bar during the 15 minutes is recorded. The resulting data are as follows:What is the alternative hypothesis? In this case, assume a nondirectional hypothesis is appropriate because there is insufficient empirical basis to warrant a directional hypothesis. What is the null hypothesis? Using ? = 0.052 tail, what is your conclusion? What error might you be making by your conclusion in part c?
Subject Bar 1 Bar 2
1 20 40
2 18 25
3 24 38
4 14 27
5 5 31
6 26 21
7 15 32
8 29 38
9 15 25
10 9 18

Answer:

It is statistically significant that mean of difference scores is different from 0

Step-by-step explanation:

Subject    Bar 1      Bar 2      Difference

1                 20          40            20

2                 18          25             7

3                 24          38            14

4                 14           27            13

5                 5            31             29

6                 26          21             -5

7                 15           32            17

8                 29          38            9

9                 15           25           10

10                9             18           9

Mean of difference =[tex]\frac{Sum}{10}=\frac{123}{10}=12.3[/tex]

Standard deviation s =[tex]\sqrt{\frac{(x_i-\bar{x})^2}{n}}=8.93[/tex]

Null Hypothesis : mean of difference scores is 0 : [tex]H_0:\mu=0[/tex]

Alternate Hypothesis : mean of difference scores is different from 0 : [tex]H_a:\mu \neq 0[/tex]

We will use t test for unpaired

[tex]t = \frac{x-\mu}{\sqrt{\frac{s^2}{n-1}}}\\t = \frac{12.3-0}{\sqrt{\frac{8.93^2}{10-1}}}\\t=4.13[/tex]

Level of significance (two tailed) =[tex]\alpha = 0.05[/tex]

Degree of freedom = n-1 = 10-1 =9

Using calculator

The p-value is .002559.

p value < α

So ,we reject null hypothesis

So, It is statistically significant that mean of difference scores is different from 0

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