Math Problem Statement

A pizza delivery chain advertises that it will deliver your pizza in 2525 minutes from when the order is placed. Being a skeptic, you decide to test and see if the mean delivery time is actually more than 2525 minutes. For the simple random sample of 1212 customers who record the amount of time it takes for each of their pizzas to be delivered, the mean is 27.627.6 minutes with a standard deviation of 4.24.2 minutes. Assume that the population distribution is approximately normal. Perform a hypothesis test using a 0.050.05 level of significance.

Step 3 of 3 :

Draw a conclusion and interpret the decision.

Solution

Let's go through the hypothesis testing process step by step to draw a conclusion and interpret the decision.

Given Data:

  • Sample size (nn) = 12
  • Sample mean (xˉ\bar{x}) = 27.6 minutes
  • Sample standard deviation (ss) = 4.2 minutes
  • Hypothesized population mean (μ0\mu_0) = 25 minutes
  • Significance level (α\alpha) = 0.05

Step 1: State the Hypotheses

We are conducting a one-tailed hypothesis test to determine if the mean delivery time is greater than 25 minutes.

  • Null Hypothesis (H0H_0): μ=25\mu = 25 (the mean delivery time is 25 minutes)
  • Alternative Hypothesis (HaH_a): μ>25\mu > 25 (the mean delivery time is greater than 25 minutes)

Step 2: Calculate the Test Statistic

We will use the t-test because the sample size is small (n=12n = 12) and the population standard deviation is unknown.

The formula for the t-test statistic is:

t=xˉμ0snt = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}

Where:

  • xˉ=27.6\bar{x} = 27.6 (sample mean)
  • μ0=25\mu_0 = 25 (hypothesized mean)
  • s=4.2s = 4.2 (sample standard deviation)
  • n=12n = 12 (sample size)

Substituting the values:

t=27.6254.212=2.61.2132.14t = \frac{27.6 - 25}{\frac{4.2}{\sqrt{12}}} = \frac{2.6}{1.213} \approx 2.14

Step 3: Determine the Critical Value

Since the test is one-tailed with a significance level of 0.050.05, we need to find the critical value of tt for n1=121=11n - 1 = 12 - 1 = 11 degrees of freedom from the t-distribution table.

For df=11df = 11 and α=0.05\alpha = 0.05, the critical value is approximately 1.796.

Step 4: Make a Decision

Now, compare the calculated test statistic t=2.14t = 2.14 with the critical value t0.05,11=1.796t_{0.05, 11} = 1.796:

  • Since 2.14>1.7962.14 > 1.796, we reject the null hypothesis.

Step 5: Conclusion and Interpretation

The test statistic is greater than the critical value, so we reject the null hypothesis. This means that, at the 0.05 significance level, there is sufficient evidence to conclude that the mean delivery time is significantly greater than 25 minutes.

Relative Questions:

  1. What if the standard deviation had been larger? How would that affect the test result?
  2. What if the sample size were larger? How would the test statistic change?
  3. What happens if we change the significance level from 0.05 to 0.01? Would the decision remain the same?
  4. How does the shape of the t-distribution change as the sample size increases?
  5. If we conducted a two-tailed test instead, what would the null and alternative hypotheses be?

Tip: When performing hypothesis testing, always consider whether the test should be one-tailed or two-tailed based on the research question.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Statistics
t-Test
Sampling Distribution

Formulas

t = (x̄ - μ₀) / (s / √n)

Theorems

Central Limit Theorem
t-distribution

Suitable Grade Level

Undergraduate Statistics