Math Problem Statement

Solve

A home improvement company is interested in whether the customer satisfaction rate is different from the 76% currently claimed. The company sponsored a survey of 268 customers and found that 212 customers were satisfied. What is the test statistic z? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is different from the claim by the company at a significance level of α = 0.01?

Solution

To solve this hypothesis testing problem, let's break it down step by step:

Problem Setup

The company claims the satisfaction rate is p0=0.76p_0 = 0.76. A survey is conducted with:

  • n=268n = 268 (sample size),
  • x=212x = 212 (number of satisfied customers).

The sample proportion p^\hat{p} is: p^=xn=212268\hat{p} = \frac{x}{n} = \frac{212}{268}

We are testing:

  • Null hypothesis H0H_0: p=0.76p = 0.76
  • Alternative hypothesis H1H_1: p0.76p \neq 0.76 (two-tailed test)

Step 1: Calculate the Test Statistic (zz)

The formula for the test statistic zz is: z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}

Step 1.1: Calculate p^\hat{p}

p^=2122680.791\hat{p} = \frac{212}{268} \approx 0.791

Step 1.2: Calculate the standard error

SE=p0(1p0)n=0.76(10.76)268=0.76×0.242680.0267SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.76 (1 - 0.76)}{268}} = \sqrt{\frac{0.76 \times 0.24}{268}} \approx 0.0267

Step 1.3: Compute zz

z=p^p0SE=0.7910.760.02671.16z = \frac{\hat{p} - p_0}{SE} = \frac{0.791 - 0.76}{0.0267} \approx 1.16

So, the test statistic is: z1.16z \approx 1.16

Step 2: Calculate the p-value

This is a two-tailed test, so we need the cumulative probability of z=1.16z = 1.16 under the standard normal curve. Using a z-table or calculator: P(Z>1.16)0.123P(Z > 1.16) \approx 0.123 Since this is two-tailed, multiply by 2: p-value=2×0.1230.246p\text{-value} = 2 \times 0.123 \approx 0.246

Step 3: Make a Conclusion

The significance level is α=0.01\alpha = 0.01. Since the p-value (0.2460.246) is greater than 0.010.01, we fail to reject the null hypothesis. There is insufficient evidence to conclude that the satisfaction rate is different from 76%76\%.


Final Answers:

  1. Test statistic zz: 1.161.16
  2. p-value: 0.2460.246
  3. Conclusion: There is not enough evidence to reject H0H_0 at α=0.01\alpha = 0.01.

Would you like further clarification on any step?
Here are 5 related questions to expand on this topic:

  1. How is the p-value interpreted in hypothesis testing?
  2. Why do we multiply the tail probability by 2 for a two-tailed test?
  3. What happens if the significance level α\alpha is changed to 0.050.05?
  4. How is the sample size nn related to the standard error?
  5. What are the implications of failing to reject the null hypothesis?

Tip: Always check whether your hypothesis test is one-tailed or two-tailed before calculating the p-value!

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Population Proportion
Two-Tailed Test
Standard Normal Distribution

Formulas

z = (p_hat - p_0) / sqrt((p_0 * (1 - p_0)) / n)
p-value for a two-tailed test = 2 * P(Z > |z|)

Theorems

Central Limit Theorem
Standard Error of a Proportion

Suitable Grade Level

Grades 11-12, College Introductory Statistics