Questions: In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 130 randomly selected students on your campus and find that 73 of them would prefer a boy. (a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 130 students, at least 73 would prefer a boy, assuming the true percentage is 45%. (b) Does this result contradict the poll? Explain. (a) The probability that at least 73 students would prefer a boy is . (Round to four decimal places as needed.)

In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 130 randomly selected students on your campus and find that 73 of them would prefer a boy.

(a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 130 students, at least 73 would prefer a boy, assuming the true percentage is 45%.

(b) Does this result contradict the poll? Explain.

(a) The probability that at least 73 students would prefer a boy is .
(Round to four decimal places as needed.)

Transcript text: In a recent poll, $45 \%$ of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 130 randomly selected students on your campus and find that 73 of them would prefer a boy. (a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 130 students, at least 73 would prefer a boy, assuming the true percentage is $45 \%$. (b) Does this result contradict the poll? Explain. (a) The probability that at least 73 students would prefer a boy is $\square$ $\square$. (Round to four decimal places as needed.)

Solution

Solution Steps

Step 1: Calculate the Probability

To approximate the probability that at least 73 students would prefer a boy, we first calculate the mean and standard deviation for the normal approximation of the binomial distribution:

\[ \text{mean} = n \cdot p = 130 \cdot 0.45 = 58.5 \]

\[ \text{std\_dev} = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{130 \cdot 0.45 \cdot 0.55} \approx 4.5 \]

Using the continuity correction, we find $ P(X \geq 73) $ as follows:

\[ P(X \geq 73) = 1 - P(X \leq 72.5) \]

The calculated probability is:

\[ P(X \geq 73) \approx 0.0068 \]

Step 2: Conduct the Hypothesis Test

We perform a hypothesis test to determine if the sample proportion is significantly different from the hypothesized population proportion $ p_0 = 0.45 $. The test statistic $ Z $ is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Substituting the values:

\[ Z = \frac{\frac{73}{130} - 0.45}{\sqrt{\frac{0.45 \cdot 0.55}{130}}} \approx 2.5563 \]

Step 3: Determine the P-value and Critical Region

The P-value associated with the test statistic is calculated as:

\[ \text{P-value} \approx 0.0053 \]

The critical region for a one-tailed test at a significance level of $ \alpha = 0.05 $ is defined as:

\[ Z > 1.6449 \]

Step 4: Conclusion

Since the P-value $ 0.0053 $ is less than the significance level $ 0.05 $, we reject the null hypothesis. This indicates that the sample proportion is significantly higher than $ 0.45 $. Therefore, the result contradicts the poll, as the sample proportion of students preferring a boy is significantly greater than the reported $ 45\% $.