Math Problem Statement

A Statistics professor has observed that for several years students score an average of 115 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. In the trial course that used this software, 235 students scored an average of 118 points on the final with a standard deviation of 8.6 points. Complete parts a) and b) below: a) Should the professor spend the money for this software? Support this recommendation with an appropriate test. Use α = 0.05. What are the null and alternative hypotheses? What is the test statistic?

Solution

To answer the question, let's go through each part step-by-step.

Problem Overview

The professor wants to determine if a new software program will increase students' scores. Previous scores (without the software) averaged at 115 points out of 150. In a trial with the software, 235 students scored an average of 118 points with a standard deviation of 8.6 points.

Part a) Hypothesis Testing

We will perform a hypothesis test to see if the observed increase in scores is statistically significant.

Step 1: Define the Null and Alternative Hypotheses

Let:

  • μ0=115\mu_0 = 115: The historical mean score without the software.

The hypotheses are:

  • Null Hypothesis (H₀): The software does not increase scores. H0:μ=115H_0: \mu = 115
  • Alternative Hypothesis (Hₐ): The software increases scores. HA:μ>115H_A: \mu > 115

This is a one-tailed test since we are only interested in whether the software increases the scores, not if it decreases them.

Step 2: Test Statistic

Since we know the sample mean, population mean, sample size, and sample standard deviation, we can use a t-test to compute the test statistic.

The formula for the test statistic tt is: t=Xˉμ0snt = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}} where:

  • Xˉ=118\bar{X} = 118: sample mean,
  • μ0=115\mu_0 = 115: population mean under H0H_0,
  • s=8.6s = 8.6: sample standard deviation,
  • n=235n = 235: sample size.

Let's substitute these values: t=1181158.6235t = \frac{118 - 115}{\frac{8.6}{\sqrt{235}}}

First, calculate the denominator: 8.62358.615.330.561\frac{8.6}{\sqrt{235}} \approx \frac{8.6}{15.33} \approx 0.561

Then, calculate tt: t=30.5615.35t = \frac{3}{0.561} \approx 5.35

Step 3: Compare with Critical Value

Using a significance level of α=0.05\alpha = 0.05, we look up the critical tt-value for a one-tailed test with n1=234n - 1 = 234 degrees of freedom. For large nn, the critical value for α=0.05\alpha = 0.05 in a one-tailed test is approximately 1.645.

Since our calculated tt-value of 5.35 is much greater than 1.645, we reject the null hypothesis.

Conclusion

There is enough statistical evidence to suggest that the software does improve students' scores. The professor should consider spending money on the software.


Would you like more details or explanations on any step? Here are five related questions that can deepen your understanding:

  1. How would the conclusion change if the sample mean was lower than 115?
  2. What would the test look like if we used a two-tailed hypothesis?
  3. How do changes in the sample size affect the test statistic in hypothesis testing?
  4. What is the interpretation of a p-value in this context?
  5. What are the implications of Type I and Type II errors in this decision?

Tip: Always check if the sample size is large enough for the Central Limit Theorem to apply when performing hypothesis testing with sample means.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
One-Sample t-test
Significance Level
Test Statistic

Formulas

t = (X̄ - μ) / (s / √n)

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate (Introductory Statistics)