Questions: Many high school students take the AP tests in different subject areas. In one year, of the 210,737 students who took the calculus AB exam 104,919 of them were female and 105,818 of them were male. Is there enough evidence to show that the proportion of female students taking the calculus AB exam is different from the proportion of male students taking the calculus AB exam? Test at the 10% level. State the hypotheses. H0: p1 = p2 Ha: p1 ≠ p2 Calculate the test statistics. Round to four decimal places. p̂1=0.4978 p̂2=0.5021 Calculate the standardized test statistic. Round three decimal places. z= Find the p-value. Round to four decimal places. p-value =

Many high school students take the AP tests in different subject areas. In one year, of the 210,737 students who took the calculus AB exam 104,919 of them were female and 105,818 of them were male. Is there enough evidence to show that the proportion of female students taking the calculus AB exam is different from the proportion of male students taking the calculus AB exam? Test at the 10% level.
State the hypotheses.

H0: p1 = p2
Ha: p1 ≠ p2

Calculate the test statistics. Round to four decimal places.

p̂1=0.4978
p̂2=0.5021

Calculate the standardized test statistic. Round three decimal places.

z=

Find the p-value. Round to four decimal places.

p-value =
Transcript text: Many high school students take the AP tests in different subject areas. In one year, of the 210,737 students who took the calculus AB exam 104,919 of them were female and 105,818 of them were male. Is there enough evidence to show that the proportion of female students taking the calculus $A B$ exam is different from the proportion of male students taking the calculus AB exam? Test at the $10 \%$ level. State the hypotheses. \[ \begin{array}{l} H_{0}: p_{1} = p_{2} \\ H_{a}: p_{1} \neq p_{2} \end{array} \] Calculate the test statistics. Round to four decimal places. \[ \begin{array}{l} \hat{p}_{1}=0.4978 \\ \hat{p}_{2}=0.5021 \end{array} \] Calculate the standardized test statistic. Round three decimal places. \[ z= \] Find the p-value. Round to four decimal places. p -value $=$ $\square$
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Solution

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Solution Steps

To solve this problem, we need to perform a hypothesis test for the difference in proportions. The null hypothesis states that the proportion of female students taking the calculus AB exam is equal to the proportion of male students. The alternative hypothesis states that these proportions are different. We will calculate the test statistic using the formula for the difference in sample proportions and then find the p-value using the standard normal distribution.

Step 1: State the Hypotheses

We are testing the following hypotheses: \[ H_{0}: p_{1} = p_{2} \quad \text{(the proportions are equal)} \] \[ H_{a}: p_{1} \neq p_{2} \quad \text{(the proportions are different)} \]

Step 2: Calculate the Sample Proportions

The sample proportions for female and male students are calculated as follows: \[ \hat{p}_{1} = \frac{n_{1}}{N} = \frac{104919}{210737} \approx 0.4979 \] \[ \hat{p}_{2} = \frac{n_{2}}{N} = \frac{105818}{210737} \approx 0.5021 \]

Step 3: Calculate the Combined Sample Proportion and Standard Error

The combined sample proportion is: \[ \hat{p} = \frac{n_{1} + n_{2}}{N} = \frac{104919 + 105818}{210737} = 0.5 \] The standard error (SE) is calculated as: \[ SE = \sqrt{\hat{p} \left(1 - \hat{p}\right) \left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)} = \sqrt{0.5 \cdot 0.5 \left(\frac{1}{104919} + \frac{1}{105818}\right)} \approx 0.0022 \]

Step 4: Calculate the Test Statistic

The standardized test statistic \( z \) is calculated using the formula: \[ z = \frac{\hat{p}_{1} - \hat{p}_{2}}{SE} = \frac{0.4979 - 0.5021}{0.0022} \approx -1.958 \]

Step 5: Calculate the p-value

The p-value for a two-tailed test is calculated as: \[ p\text{-value} = 2 \cdot \left(1 - \Phi(|z|)\right) \approx 0.0502 \] where \( \Phi \) is the cumulative distribution function of the standard normal distribution.

Final Answer

The standardized test statistic is \( z \approx -1.958 \) and the p-value is \( \approx 0.0502 \).

Thus, we have: \[ \boxed{z = -1.958} \] \[ \boxed{p\text{-value} = 0.0502} \]

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