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

Answer:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

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

Step-by-step 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:

$t =\frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we got:

$t = \frac{39.5-30.6}{\frac{6.6}{\sqrt{25}}}= 6.74$

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