Question

Five students were tested before and after taking a class to improve their study habits. They were given articles to read which contained a known number of facts in each story. After the story each student listed as many facts as he/she could recall. The following data was recorded.

Before 10 12 14 16 12
After 15 14 17 17 20

The obtained value of the appropriate statistic is ____.

a. 3.92
b. 2.58
c. 4.12
d. 3.06

What do you conclude using a = 0.052 tail?

a. retain H0; we cannot conclude that the class improved study habits
b. accept H0; the class appeared to improve study habits
c. retain H0; the class had no effect on study habits
d. reject H0; the class appeared to improve study habits

Answer 1

Answer:

The obtained value of the appropriate statistic is ____.

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06$  

d. 3.06

$p_v =2*P(t_{(4)}>3.06) =0.0376$  

d. reject H0; the class appeared to improve study habits

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations we can use it.  

Let put some notation  

x=before method , y = after method  

x: 10 12 14 16 12

y: 15 14 17 17 20

The system of hypothesis for this case are:  

Null hypothesis: $\mu_y- \mu_x = 0$  

Alternative hypothesis: $\mu_y -\mu_x \neq 0$  

The first step is define the difference $d_i=y_i-x_i$, that is given so we have:

d: 5,2,3,1,8

The second step is calculate the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=3.8$  

The third step would be calculate the standard deviation for the differences, and we got:  

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =2.775$  

The fourth step is calculate the statistic given by :  

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{3.8 -0}{\frac{2.775}{\sqrt{5}}}=3.06$  

The next step is calculate the degrees of freedom given by:  

$df=n-1=5-1=4$  

Now we can calculate the p value, since we have a two tailed test the p value is given by:  

$p_v =2*P(t_{(4)}>3.06) =0.0376$  

The p value is less than the significance level given $\alpha=0.05$, so then we can conclude that we reject the null hypothesis.

d. reject H0; the class appeared to improve study habits