Math Problem Statement

A marketing manager for a cell phone company claims that less than 59% of children aged 8 - 12 have cell phones. In a survey of 822 children aged 8 - 12 by a national consumer group, 460 of them had cell phones. Can you conclude that the manager's claim is true? Use the =α0.01 level of significance and the critical value method with the table. State the appropriate null and alternate hypotheses.

H0: H1: This hypothesis test is a (left or right-tailed or two-tailed) test.

Solution

To determine if the marketing manager's claim is valid, we need to conduct a hypothesis test for a population proportion.

Step 1: State the Hypotheses

The null and alternate hypotheses are structured as follows:

  • H0H_0 (Null hypothesis): The proportion of children aged 8–12 with cell phones is greater than or equal to 59%.
    • H0:p0.59H_0: p \geq 0.59
  • H1H_1 (Alternate hypothesis): The proportion of children aged 8–12 with cell phones is less than 59%.
    • H1:p<0.59H_1: p < 0.59

This is a left-tailed test because the claim suggests that the proportion is less than 59%.

Step 2: Significance Level

The significance level is given as α=0.01\alpha = 0.01.

Step 3: Sample Proportion

The sample size n=822n = 822, and the number of children with cell phones x=460x = 460.

The sample proportion p^\hat{p} is calculated as: p^=xn=4608220.5595\hat{p} = \frac{x}{n} = \frac{460}{822} \approx 0.5595

Step 4: Test Statistic

The test statistic for a population proportion is calculated as: z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} Where:

  • p^\hat{p} is the sample proportion
  • p0=0.59p_0 = 0.59 is the hypothesized population proportion
  • n=822n = 822 is the sample size

Substituting in the values: z=0.55950.590.59(10.59)822z = \frac{0.5595 - 0.59}{\sqrt{\frac{0.59(1 - 0.59)}{822}}} z=0.03050.59×0.41822=0.03050.0002940.03050.017141.78z = \frac{-0.0305}{\sqrt{\frac{0.59 \times 0.41}{822}}} = \frac{-0.0305}{\sqrt{0.000294}} \approx \frac{-0.0305}{0.01714} \approx -1.78

Step 5: Critical Value

Since this is a left-tailed test and α=0.01\alpha = 0.01, we find the critical value from the standard normal distribution table (z-table). The critical value for α=0.01\alpha = 0.01 in a left-tailed test is -2.33.

Step 6: Conclusion

  • The test statistic z1.78z \approx -1.78 is not less than the critical value of z=2.33z = -2.33.
  • Therefore, we fail to reject the null hypothesis H0H_0.

Final Conclusion:

There is not enough evidence at the α=0.01\alpha = 0.01 level of significance to support the marketing manager's claim that less than 59% of children aged 8–12 have cell phones.


Would you like more details on any step or have questions?

Here are 5 related questions to consider:

  1. What is the significance of using a left-tailed test in this scenario?
  2. How does the choice of α=0.01\alpha = 0.01 affect the results?
  3. How would the conclusion change if the test statistic was smaller than the critical value?
  4. What happens if we choose a different level of significance, such as α=0.05\alpha = 0.05?
  5. Can we calculate a confidence interval for the sample proportion?

Tip: When conducting hypothesis tests, always double-check the direction of the test (left-tailed, right-tailed, or two-tailed) to ensure you're using the correct critical values.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Population Proportion
Significance Level
Normal Distribution

Formulas

Sample Proportion: p̂ = x/n
Test Statistic for Population Proportion: z = (p̂ - p₀) / √[p₀(1 - p₀) / n]

Theorems

Central Limit Theorem
Critical Value Method

Suitable Grade Level

College-level (Statistics, Probability)