Math Problem Statement

College tuition: The mean annual tuition and fees for a sample of

13

private colleges in California was

$38,000

with a standard deviation of

$7900

. A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California differs from

$35,000

? Use the

=α0.01

level of significance and the

P

-value method with the TI-84 Plus calculator. (c) Compute the

P

-value

Solution

To determine whether the mean tuition and fees for private colleges in California differs from $35,000, we perform a hypothesis test using the given data and compute the P-value. Here's how it’s done:


Step 1: Hypotheses

  • Null Hypothesis (H0H_0): μ=35,000\mu = 35,000 (the mean tuition is $35,000).
  • Alternative Hypothesis (HaH_a): μ35,000\mu \neq 35,000 (the mean tuition is not $35,000).

This is a two-tailed test.


Step 2: Test Statistic

We use the formula for the t-test statistic for a single sample: t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}} where:

  • xˉ=38,000\bar{x} = 38,000 (sample mean),
  • μ=35,000\mu = 35,000 (hypothesized mean),
  • s=7,900s = 7,900 (sample standard deviation),
  • n=13n = 13 (sample size).

Substituting the values: t=38,00035,0007,900/13=3,0007,900/3.605=3,0002,191.61.37t = \frac{38,000 - 35,000}{7,900 / \sqrt{13}} = \frac{3,000}{7,900 / 3.605} = \frac{3,000}{2,191.6} \approx 1.37


Step 3: Degrees of Freedom

Degrees of freedom (dfdf) for a single-sample t-test is: df=n1=131=12df = n - 1 = 13 - 1 = 12


Step 4: Compute the P-Value

Using a TI-84 calculator:

  1. Press 2ndDISTRtcdf(.
  2. Input the bounds for the t-distribution:
    • Lower bound = -\infty (use a very large negative value like 100-100),
    • Upper bound = 1.37-1.37 (for the left tail),
    • Degrees of freedom = 12.
  3. Repeat for the upper tail with t=1.37t = 1.37.
  4. Multiply the result by 2 (since it’s a two-tailed test).

Alternatively, approximate manually: Using t-distribution tables, for t=1.37t = 1.37 and df=12df = 12, the one-tailed P-value is approximately 0.1. Thus, the two-tailed P-value is: P2×0.1=0.2P \approx 2 \times 0.1 = 0.2


Step 5: Decision Rule

At α=0.01\alpha = 0.01:

  • Compare the P-value (0.2) to α\alpha.
  • Since P>αP > \alpha, fail to reject the null hypothesis.

Conclusion

There is insufficient evidence at the α=0.01\alpha = 0.01 level of significance to conclude that the mean tuition and fees for private colleges in California differs from $35,000.


Would you like further clarification or step-by-step instructions for using the TI-84?


5 Questions to Explore Further

  1. How is the t-distribution different from the standard normal distribution?
  2. What happens if the sample size increases while keeping all other variables constant?
  3. How do we determine whether to use a one-tailed or two-tailed test?
  4. What are the assumptions of the single-sample t-test?
  5. How would the result change if α=0.05\alpha = 0.05 instead of 0.01?

Tip:

Always check if the population is approximately normal or if the sample size is large (n30n \geq 30) before using t-tests to ensure valid results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
P-value Method
Two-tailed Tests
t-distribution

Formulas

t = (x̄ - μ) / (s / √n)
Degrees of Freedom (df) = n - 1
P-value from t-distribution

Theorems

Central Limit Theorem
Properties of the t-distribution

Suitable Grade Level

College-level Statistics