Question

A psychologist is interested in knowing whether adults who were bullied as children differ from the general population in terms of their empathy for others. On a questionnaire designed to measure empathy, the mean score for the general population is 30.6. Random sampling of 25 scores obtained from individuals who were bullied yielded a sample mean of 39.5 and a sample standard deviation of 6.6. Test at the .05 level of significance. Suppose that the value of your calculated (obtained) test statistic is 6.74. What is your decision

Answer 1

`t = (39.5-30.6)/((6.6)/(√(25)))= 6.74`

The degrees of freedom are given by:

`df =n-1= 25-1=24`

Now we can calculate the p value with the following probability:

`p_v = 2*P(t_(24)>6.74)= 5.69x10^(-7)`

And for this case since the p value is lower compared to the significance level
`\alpha=0.05` we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05

Explanation:

For this case we have the following info given:

`\bar X = 39.5` represent the sample mean

`s =6.6` represent the sample deviation

`\mu = 30.6` represent the reference value to test.

`n = 25` represent the sample size selected

The statistic for this case is given by:

`t =(\bar X -\mu)/((s)/(√(n)))`

And replacing we got:

`t = (39.5-30.6)/((6.6)/(√(25)))= 6.74`

The degrees of freedom are given by:

`df =n-1= 25-1=24`

Now we can calculate the p value with the following probability:

`p_v = 2*P(t_(24)>6.74)= 5.69x10^(-7)`

And for this case since the p value is lower compared to the significance level
`\alpha=0.05` we can reject the null hypothesis and we can conclude that the true mean for this case is different from 30.6 at the significance level of 0.05