Question

A tuter promises to improve GMAT scores of students by more than 50 points after three lessons. To see if this is true, the tutor takes a sample of 49 students’ test scores after and before they received tutoring. The mean difference was 53 points better after tutoring, with a standard deviation of the difference equal to 12 points. Let µD denote the mean of the difference: score after tutoring minus score before tutoring. Which of the following is the correct value of the test statistic?
t48 = 1.7143

t48 = 1.7500

z = 1.7500

z = 1.7143

Answer 1

Answer:

t48 = 1.7500

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 (This problem) we can use it.  

Let put some notation  

x=test value before , y = test value after  

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \geq 50$

Alternative hypothesis: $\mu_y -\mu_x >50$

We have the mean difference  

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=53$

And the sample standard deviation for the differences, is:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =12$

Now we can calculate the statistic given by :

$t=\frac{\bar d -\Delta}{\frac{s_d}{\sqrt{n}}}=\frac{53 -50}{\frac{12}{\sqrt{49}}}=1.75$

Since we don't have the population standard deviaition we need to use the t distribution and we can calculate the degrees of freedom given by:

$df=n-1=49-1=48$

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

$p_v =P(t_{(48)}>1.75) =0.0432$

If we use a significance level of 5% we have that $p_v$, and on this case we reject the null hypothesis . And we don't have enough evidence to conclude that the difference of means it's higher than 50 points.