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.

*H*0: *H*1: This hypothesis test is a (left or right-tailed or two-tailed) test.

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:

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

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 \( \alpha = 0.01 \).

### Step 3: Sample Proportion
The sample size \( n = 822 \), and the number of children with cell phones \( x = 460 \).

The sample proportion \( \hat{p} \) is calculated as:
\[
\hat{p} = \frac{x}{n} = \frac{460}{822} \approx 0.5595
\]

### Step 4: Test Statistic
The test statistic for a population proportion is calculated as:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \( \hat{p} \) is the sample proportion
- \( p_0 = 0.59 \) is the hypothesized population proportion
- \( n = 822 \) is the sample size

Substituting in the values:
\[
z = \frac{0.5595 - 0.59}{\sqrt{\frac{0.59(1 - 0.59)}{822}}}
\]
\[
z = \frac{-0.0305}{\sqrt{\frac{0.59 \times 0.41}{822}}} = \frac{-0.0305}{\sqrt{0.000294}} \approx \frac{-0.0305}{0.01714} \approx -1.78
\]

### Step 5: Critical Value
Since this is a left-tailed test and \( \alpha = 0.01 \), we find the critical value from the standard normal distribution table (z-table). The critical value for \( \alpha = 0.01 \) in a left-tailed test is **-2.33**.

### Step 6: Conclusion
- The test statistic \( z \approx -1.78 \) is **not less than** the critical value of \( z = -2.33 \).
- Therefore, we **fail to reject** the null hypothesis \( H_0 \).

### Final Conclusion:
There is not enough evidence at the \( \alpha = 0.01 \) 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 \( \alpha = 0.01 \) 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 \( \alpha = 0.05 \)?
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.

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 α = 0.01 level of significance and the critical value method with the table. State the appropriate null and alternate hypotheses.

Test the claim that less than 59% of children aged 8-12 have cell phones using hypothesis testing at a 0.01 significance level. This example walks through calculating sample proportion, test statistics, and determining if the null hypothesis can be rejected.

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