Math Problem Statement

In a study of cell phone usage and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 6962 subjects randomly selected from an online group involved with ears. There were 1290 surveys returned. Use a 0.01 significance level to test the claim that the return rate is less than​ 20%. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Question content area bottom Part 1 Identify the null hypothesis and alternative hypothesis. A. Upper H 0​: pequals0.2 Upper H 1​: pnot equals0.2 B. Upper H 0​: pnot equals0.2 Upper H 1​: pequals0.2 C. Upper H 0​: pgreater than0.2 Upper H 1​: pequals0.2 D. Upper H 0​: pequals0.2 Upper H 1​: pgreater than0.2 E. Upper H 0​: pequals0.2 Upper H 1​: pless than0.2 Your answer is correct.F. Upper H 0​: pless than0.2 Upper H 1​: pequals0.2 Part 2 The test statistic is z =   

​(Round to two decimal places as​ needed.)

Solution

Part 1: Hypothesis Identification

We are testing whether the return rate of surveys is less than 20%. This indicates that the null hypothesis assumes the return rate is 20%, and the alternative hypothesis suggests that the return rate is less than 20%.

  • Null Hypothesis (H0H_0): The proportion of returned surveys is 20% (i.e., p=0.20p = 0.20).
  • Alternative Hypothesis (H1H_1): The proportion of returned surveys is less than 20% (i.e., p<0.20p < 0.20).

Thus, the correct hypothesis pair is:

E. H0:p=0.20H_0: p = 0.20, H1:p<0.20H_1: p < 0.20

Part 2: Calculating the Test Statistic

We will use the normal approximation to the binomial distribution to compute the test statistic.

Step 1: Identify the values

  • Sample size (nn) = 1290 (the number of surveys returned)
  • Sample proportion (p^\hat{p}) = 12906962=0.185\frac{1290}{6962} = 0.185 (the proportion of returned surveys)
  • Hypothesized proportion (p0p_0) = 0.20

Step 2: Compute the standard error of the proportion

The standard error (SESE) for the sample proportion is calculated using the formula: SE=p0(1p0)nSE = \sqrt{\frac{p_0(1 - p_0)}{n}} Substitute p0=0.20p_0 = 0.20 and n=6962n = 6962: SE=0.20(10.20)6962=0.20(0.80)6962=0.1669620.000022950.00479SE = \sqrt{\frac{0.20(1 - 0.20)}{6962}} = \sqrt{\frac{0.20(0.80)}{6962}} = \sqrt{\frac{0.16}{6962}} \approx \sqrt{0.00002295} \approx 0.00479

Step 3: Calculate the test statistic (z)

The z-test statistic is calculated by the formula: z=p^p0SEz = \frac{\hat{p} - p_0}{SE} Substitute p^=0.185\hat{p} = 0.185, p0=0.20p_0 = 0.20, and SE=0.00479SE = 0.00479: z=0.1850.200.00479=0.0150.004793.14z = \frac{0.185 - 0.20}{0.00479} = \frac{-0.015}{0.00479} \approx -3.14

So, the test statistic is z ≈ -3.14.

Summary of Results:

  • Null Hypothesis (H0H_0): p=0.20p = 0.20
  • Alternative Hypothesis (H1H_1): p<0.20p < 0.20
  • Test Statistic (z): -3.14

Would you like to continue with calculating the p-value or testing the hypothesis further? Let me know if you'd like additional steps or clarifications.


Related Questions:

  1. How do you compute the p-value for a given z-score?
  2. What is the interpretation of the p-value in hypothesis testing?
  3. How would the test change if the significance level was 0.05 instead of 0.01?
  4. What does the normal approximation to the binomial distribution imply in this context?
  5. How do you interpret the results if the z-score were greater than the critical value?

Tip: In hypothesis testing, always ensure your sample size is large enough to use the normal approximation, and check if both np010np_0 \geq 10 and n(1p0)10n(1 - p_0) \geq 10.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Normal Distribution Approximation
Z-Tests

Formulas

Standard Error (SE) = sqrt(p0 * (1 - p0) / n)
Test Statistic (z) = (p̂ - p0) / SE

Theorems

Central Limit Theorem (normal approximation to binomial distribution)

Suitable Grade Level

Grades 11-12