 A marketing manager for a phone company claims that more than 54% of children aged 8-12 have phones. In a survey of 812 children aged 8-12 by a national consumers group,  of them had phones. Can you conclude that the manager's claim is true? Use the a=0.05  level of significance and the P-value method with the TI-84 Plus calculator.

To determine whether more than 54% of children aged 8-12 have phones, given the sample data and the significance level, we can perform a one-proportion z-test. Here are the steps:

### Step 1: State the Hypotheses
- Null Hypothesis (\(H_0\)): \(p = 0.54\)
- Alternative Hypothesis (\(H_a\)): \(p > 0.54\)

### Step 2: Calculate the Test Statistic
Let's denote:
- Sample size (\(n\)) = 812
- Number of children with phones (\(x\)) = \( \) (let's call it \( x \) to denote the value we'll insert)
- Sample proportion (\(\hat{p}\)) = \(\frac{x}{812}\)

The test statistic for a one-proportion z-test is calculated using:
\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Where:
- \(\hat{p}\) is the sample proportion
- \(p_0\) is the population proportion under the null hypothesis (0.54)
- \(n\) is the sample size

Let's calculate it assuming that \( x \) was given in the survey results.

### Step 3: Calculate the P-value
Using the z-table or the TI-84 Plus calculator, we can find the P-value corresponding to the calculated z-value.

On the TI-84 Plus:
1. Press `2nd` and then `VARS` to access the distribution menu.
2. Choose `normalcdf`.
3. Enter the lower bound (the z-value), the upper bound (1E99 since we are interested in the area to the right), and the mean and standard deviation (0 and 1 for the standard normal distribution).

The P-value will be calculated and displayed.

### Step 4: Compare the P-value with the Significance Level
- At \(\alpha = 0.05\):
  - If the P-value is less than 0.05, we reject the null hypothesis.
  - If the P-value is greater than or equal to 0.05, we do not reject the null hypothesis.

### Conclusion
Based on the comparison, we can conclude whether there is sufficient evidence to support the manager's claim that more than 54% of children aged 8-12 have phones.

Let's insert a specific value for \( x \) (number of children with phones) to illustrate the calculation. Assume \( x = 500 \).

Given \( x = 500 \):
- Sample proportion (\(\hat{p}\)) = \(\frac{500}{812} \approx 0.615\)

Calculate the z-value:
\[ z = \frac{0.615 - 0.54}{\sqrt{\frac{0.54(1 - 0.54)}{812}}} \approx \frac{0.075}{\sqrt{\frac{0.54 \cdot 0.46}{812}}} \approx \frac{0.075}{0.0175} \approx 4.29 \]

Calculate the P-value:
- Using the TI-84 Plus: `normalcdf(4.29, 1E99, 0, 1)` gives a very small P-value, much less than 0.05.

Since the P-value is much smaller than 0.05, we reject the null hypothesis.

### Conclusion
There is sufficient evidence at the \(\alpha = 0.05\) level of significance to support the manager's claim that more than 54% of children aged 8-12 have phones.

Would you like further details or have any questions?

Here are five questions you might find interesting:
1. How does the choice of significance level (\(\alpha\)) affect the hypothesis test results?
2. What is the impact of sample size on the margin of error in hypothesis testing?
3. Can you explain Type I and Type II errors in the context of hypothesis testing?
4. How do confidence intervals relate to hypothesis tests?
5. What are the assumptions required for performing a one-proportion z-test?

**Tip:** Always double-check the assumptions and conditions for a hypothesis test to ensure the validity of the test results.

Explore whether more than 54% of children aged 8-12 have phones using hypothesis testing. Step-by-step guide with calculations, significance level details, and P-value method on TI-84 Plus calculator. Data from a survey of 812 children aged 8-12 by a national consumers group.

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