Questions: A national health survey weighed a sample of 270 boys aged 6-11 and found that 40 of them were overweight. They weighed a sample of 220 girls aged 6-11 and found that 32 of them were overweight. Can you conclude that the percentage of overweight kids aged 6-11 among boys is greater than the percentage of overweight kids aged 6-11 among girls? Use level of significance 10%.

A national health survey weighed a sample of 270 boys aged 6-11 and found that 40 of them were overweight. They weighed a sample of 220 girls aged 6-11 and found that 32 of them were overweight. Can you conclude that the percentage of overweight kids aged 6-11 among boys is greater than the percentage of overweight kids aged 6-11 among girls? Use level of significance 10%.

Transcript text: A national health survey weighed a sample of 270 boys aged $6-11$ and found that 40 of them were overweight. They weighed a sample of 220 girls aged 6-11 and found that 32 of them were overweight. Can you conclude that the percentage of overweight kids aged 6-11 among boys is greater than the percentage of overweight kids aged 6-11 among girls? Use level of significance 10\%. Procedure: Select'an answer Assumptions: (select everything that applies) The number of positive and negative responses are both greater than 10 for both samples Simple random samples Sample sizes are both greater than 30 Population standard deviation are unknown but assumed equal Normal populations Independent samples Population standard deviations are known Population standard deviation are unknown Paired samples

Solution

Solution Steps

Step 1: Define the Problem

We want to determine if the percentage of overweight boys aged $6-11$ is greater than the percentage of overweight girls aged $6-11$. We have the following data:

Sample size for boys ($n_1$): $270$
Number of overweight boys ($x_1$): $40$
Sample size for girls ($n_2$): $220$
Number of overweight girls ($x_2$): $32$

Step 2: Calculate Sample Proportions

The sample proportions for boys and girls are calculated as follows: \[ p_1 = \frac{x_1}{n_1} = \frac{40}{270} \approx 0.1481 \] \[ p_2 = \frac{x_2}{n_2} = \frac{32}{220} \approx 0.1455 \]

Step 3: Calculate the Pooled Proportion

The pooled proportion is calculated using the formula: \[ p_{\text{pool}} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{40 + 32}{270 + 220} = \frac{72}{490} \approx 0.1469 \]

Step 4: Calculate the Standard Error

The standard error (SE) for the difference in proportions is given by: \[ SE = \sqrt{p_{\text{pool}} \cdot (1 - p_{\text{pool}}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)} = \sqrt{0.1469 \cdot (1 - 0.1469) \left( \frac{1}{270} + \frac{1}{220} \right)} \approx 0.0322 \]

Step 5: Calculate the Z-Statistic

The z-statistic is calculated as: \[ z = \frac{p_1 - p_2}{SE} = \frac{0.1481 - 0.1455}{0.0322} \approx 0.0838 \]

Step 6: Determine the Critical Value and P-Value

For a right-tailed test at a significance level of $ \alpha = 0.10 $, the critical value is: \[ z_{\text{critical}} \approx 1.2816 \] The p-value associated with the z-statistic is: \[ p\text{-value} \approx 0.4666 \]

Step 7: Conclusion

Since the calculated z-statistic $0.0838$ is less than the critical value $1.2816$ and the p-value $0.4666$ is greater than $0.10$, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the percentage of overweight boys is greater than the percentage of overweight girls.

Final Answer

$\boxed{\text{Fail to reject the null hypothesis: There is not enough evidence to conclude that the percentage of overweight boys is greater than the percentage of overweight girls.}}$

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