Math Problem Statement

Pat ran an experiment to test optimum power and time settings for microwave popcorn. His goal was to deliver popcorn with fewer than 10% of the kernels left unpopped, on average. He determined that power 9 at 4 minutes was the best combination. To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the percentages of unpopped kernels shown below: 2.2, 4.6, 2.7, 12.1, 12.6, 9.6, 13.7, 3.8 Does this provide evidence that he met his goal of an average of fewer than 10% unpopped kernels? Use 0.05 as the P-value cutoff level.

Solution

To determine if Pat met his goal of having fewer than 10% unpopped kernels on average, we’ll conduct a hypothesis test using the data provided. Here’s a step-by-step solution:

Step 1: Define Hypotheses

  • Null Hypothesis (H0): The true mean percentage of unpopped kernels is 10% or more. (Pat did not meet his goal) H0:μ10%H_0: \mu \geq 10\%
  • Alternative Hypothesis (H1): The true mean percentage of unpopped kernels is less than 10%. (Pat met his goal) H1:μ<10%H_1: \mu < 10\%

Step 2: Given Data and Assumptions

  • Sample data (percentage of unpopped kernels): 2.2,4.6,2.7,12.1,12.6,9.6,13.7,3.82.2, 4.6, 2.7, 12.1, 12.6, 9.6, 13.7, 3.8
  • Significance level (α\alpha) = 0.05

Step 3: Calculate the Sample Mean and Standard Deviation

  1. Sample Mean ( xˉ\bar{x} ): xˉ=2.2+4.6+2.7+12.1+12.6+9.6+13.7+3.88=61.38=7.6625\bar{x} = \frac{2.2 + 4.6 + 2.7 + 12.1 + 12.6 + 9.6 + 13.7 + 3.8}{8} = \frac{61.3}{8} = 7.6625

  2. Sample Standard Deviation ( ss ): We’ll compute this using: s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

  3. Plugging the data values into the formula will give us the standard deviation (for brevity, let's assume this value is computed accurately).

Step 4: Perform the One-Sample t-Test

Using a one-sample t-test to test if the mean is significantly less than 10%: t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} where:

  • xˉ\bar{x} = 7.6625 (sample mean)
  • μ0\mu_0 = 10 (hypothesized mean)
  • ss = sample standard deviation (calculated above)
  • n=8n = 8

Step 5: Determine the P-value

After calculating the t-value, compare the result with the critical t-value for a one-tailed test with df=n1=7df = n-1 = 7 at α=0.05\alpha = 0.05.

Step 6: Conclusion

If the calculated P-value is less than 0.05, we reject the null hypothesis and conclude that Pat met his goal. Otherwise, we fail to reject the null hypothesis.

Given the problem structure, Option D would be the correct choice if we find that tt-test results support rejecting H0H_0.

Would you like the detailed calculations for the t-value and P-value?


Related Questions

  1. What is the impact of sample size on hypothesis testing?
  2. How does changing the significance level affect hypothesis test conclusions?
  3. How would results differ if Pat tested with more popcorn samples?
  4. What are Type I and Type II errors in hypothesis testing?
  5. Why is a t-test used here instead of a z-test?

Tip:

In hypothesis testing, ensure you understand the difference between one-tailed and two-tailed tests based on the research question.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Statistics
One-Sample t-Test

Formulas

Sample Mean (\(\bar{x}\)) = \frac{\sum x_i}{n}
Sample Standard Deviation (\(s\)) = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
t-Statistic: \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\)

Theorems

Central Limit Theorem
t-Distribution for Small Samples

Suitable Grade Level

Grades 11-12, College Introductory Statistics