Question

When I sit and watch my students take exams, I often think to myself "I wonder if students with bright calculators are impacted by the pretty colors." This leads me to wonder if there is a difference in the exam scores of students with colorful calculators versus the exam scores of students with plain black calculators. To investigate further, I took a sample of past students. There were 49 students with colorful calculators with a mean exam score of 84 and a corresponding standard deviation of 4.7. There were 38 students with plain black calculators with a mean exam score of 87 and a corresponding standard deviation of 5.7. Is there sufficient evidence to suggest the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators

Answer 1

Answer:

There is a difference between the two means.

Step-by-step explanation:

The hypothesis can be defined as:

H₀: The mean exam scores of my SAT 215 students with colorful calculators are same as the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ = 0.

Hₐ: The mean exam scores of my SAT 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators, i.e. μ₁ - μ₂ ≠ 0.

Assume that the significance level of the test is, α = 0.05. Also assuming that the population variances are equal.

The decision rule:

A 95% confidence interval for mean difference can be used to determine the result of the hypothesis test. If the 95% confidence interval contains the null hypothesis value, i.e. 0 then the null hypothesis will not be rejected.

The 95% confidence interval for mean difference is:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

Compute the pooled standard deviation as follows:

$S_{p}=\sqrt{\frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}} {n_{1}+n_{2}-2}}}=\sqrt{\frac{(49-1)(4.7)^{2}+(38-1)(5.7)^{2}}{49+38-2}}=5.16$

The critical value of t is:

$t_{\alpha/2, (n_{1}+n_{2}-2)}=t_{0.05/2, (49+38-2)}=t_{0.025, 85}=1.984$

*Use a t-table.

Compute the 95% confidence interval for mean difference as follows:

$CI=\bar x_{1}-\bar x_{2}\pm t_{\alpha/2, (n_{1}+n_{2}-2)}\times S_{p}\times \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}$

     $=(84-87)\pm 1.984\times 5.16\times \sqrt{\frac{1}{49}+\frac{1}{38}}$

     $=-3\pm 2.133\\=(-5.133, -0.867)\\\approx(-5.13, -0.87)$

The 95% confidence interval for mean difference is (-5.13, -0.87).

The confidence interval does not contains the value 0. This implies that the null hypothesis will be rejected at 5% level of significance.

Hence, concluding that the mean exam scores of my STA 215 students with colorful calculators are different than the mean scores of my STA 215 students with plain black calculators.