Question

A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple-choice test about the issue in which people in general (who know nothing about the issue) usually get 40 questions correct. The number correct for these five individuals was 48, 41, 40, 51, and 50. Using the .05 level of significance, two-tailed, do people who see the ads score differently on this test
Use steps of hypothesis testing sketch distribution involved

Answer 1

Answer:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$    

The degrees of freedom are given by:

$df=n-1=5-1=4$  

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=46$ represent the mean height for the sample  

$s=5.148$ represent the sample standard deviation

$n=5$ sample size  

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:$\mu = 40$  

Alternative hypothesis:$\mu \neq 40$  

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Replacing we got:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$    

The degrees of freedom are given by:

$df=n-1=5-1=4$  

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached