Question

It is believed that students who begin studying for final exams a week before the test score differently than students who wait until the night before. Suppose you want to test the hypothesis that students who study one week before score greater than students who study the night before, giving you the following hypotheses: Null Hypothesis: μ1 ≤ μ2, Alternative Hypothesis: μ1 > μ2. A random sample of 31 students who indicated they studied early shows an average score of 86.19 (SD = 4.358) and 35 randomly selected procrastinators had an average score of 84.58 (SD = 6.631). Perform a two independent samples t-test assuming that early studiers are group 1 and procrastinators are group 2. What is the test statistic and p-value of this test? Assume the population standard deviations are the same.

Answer 1

Answer:

The test statistic value is 1.15.

The p-value of the test statistic is 0.1272.

Step-by-step explanation:

A hypothesis test is performed to determine whether the students who study one week before score greater than students who study the night before.

The students in the group who study early for the exam is marked as 1 and the students in the group who study a night before the exam are marked as 2.

The hypothesis is defined as:

H₀: The students who study one week before does not score greater than students who study the night before, i.e. μ₁ ≤ μ₂.

Hₐ: The students who study one week before score greater than students who study the night before, i.e. μ₁ > μ₂.

The information provided is:

$n_{1}=31\\\bar x_{1}=86.19\\s_{1}=4.358\\n_{2}=35\\\bar x_{2}=84.58\\s_{1}=6.631$

A two-sample t test is used to perform the test.

Assumption:

The two population variances are assumed to be equal.

The test statistic is given by:

$t=\frac{\bar x_{1}-\bar x_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$

Compute the value of pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2}}=\sqrt{\frac{(31-1)4.358^{2}+(35-1)6.631^{2}}{31+35-2}}=5.68$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{S_{p}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}=\frac{86.19-84.58}{5.8\times\sqrt{\frac{1}{31}+\frac{1}{35}}}=1.15$

The test statistic value is 1.15.

The p-value of the test statistic is given by:

$P(t_{n_{1}+n_{2}-2}>t)=P(t_{64}>1.15)=0.1272$

*Use a t-table for the probability.

Thus, the p-value of the test statistic is 0.1272.

Conclusion:

As the p-value of the test is quite large the null hypothesis will not be rejected at any level of significance.

Hence concluding that the students who study one week before does not score greater than students who study the night before.