the director of student services at oxnard co...
the director of student services at oxnard college is interested in whether women are more likely to attend orientation than men before they begin their coursework. a random sample of freshmen at oxnard college were asked what their gender is and whether they attended orientation. the results of the survey are shown below:\ndata for gender vs. orientation attendance\n| | women | men |\n|--|--|--|\n| yes | 349 | 334 |\n| no | 336 | 416 |\nwhat can be concluded at the $alpha = 0.01$ level of significance? if the calculator asks, be sure to use the \pooled\ data option.\nfor this study, we should use z - test for the difference between two population proportions\n\na. the null and alternative hypotheses would be:\n$h_0$: $p1$ = $p2$ (please enter a decimal and note that $p1$ and $mu1$ represent the proportion and mean for women and $p2$ and $mu2$ represent the proportion and mean for men.)\n$h_1$: $p1$ > $p2$ (please enter a decimal)\n\nb. the test statistic $z$ = (please show your answer to 3 decimal places.)\n\nc. the p - value = (please show your answer to 4 decimal places.)\n\nd. the p - value is $s$ $alpha$\n\ne. based on this, we should reject the null hypothesis.\n\nf. thus, the final conclusion is that...
Answer
# Explanation:
## Step1: Calculate sample proportions
Let $n_1$ be the number of women and $n_2$ be the number of men. $n_1=349 + 336=685$, $n_2=334 + 416 = 750$.
The sample - proportion of women who attended orientation is $\hat{p}_1=\frac{349}{685}\approx0.51$.
The sample - proportion of men who attended orientation is $\hat{p}_2=\frac{334}{750}\approx0.445$.
The pooled proportion $\hat{p}=\frac{349 + 334}{685+750}=\frac{683}{1435}\approx0.476$.
## Step2: Calculate the test statistic
The formula for the $z$ - test statistic for the difference between two population proportions is $z=\frac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1}+\frac{1}{n_2})}}$.
Substitute the values:
\[
\begin{align*}
z&=\frac{0.51 - 0.445}{\sqrt{0.476\times(1 - 0.476)\times(\frac{1}{685}+\frac{1}{750})}}\\
&=\frac{0.065}{\sqrt{0.476\times0.524\times(\frac{750 + 685}{685\times750})}}\\
&=\frac{0.065}{\sqrt{0.249424\times\frac{1435}{513750}}}\\
&=\frac{0.065}{\sqrt{0.249424\times0.0028}}\\
&=\frac{0.065}{\sqrt{0.0006983872}}\\
&=\frac{0.065}{0.0264}\\
&\approx2.462
\end{align*}
\]
## Step3: Calculate the p - value
Since this is a right - tailed test ($H_1:p_1>p_2$), the p - value is $P(Z>z)$.
$P(Z > 2.462)=1 - P(Z\leqslant2.462)$.
From the standard normal table, $P(Z\leqslant2.462)\approx0.9931$.
So the p - value is $1 - 0.9931 = 0.0069$.
## Step4: Make a decision
The significance level $\alpha = 0.01$. Since the p - value ($0.0069$) is less than $\alpha(0.01)$, we reject the null hypothesis.
# Answer:
a. $H_0:p_1 = p_2$; $H_1:p_1>p_2$
b. $2.462$
c. $0.0069$
d. $<$
e. reject
f. There is sufficient evidence at the $\alpha = 0.01$ level of significance to conclude that women are more likely to attend orientation than men at Oxnard College.