`0.67 - 0.5)/0.021 (`

`[1] 8.095238`

In a study, we find that 67% of women in a random sample view divorce as morally acceptable. Does this provide evidence that more than 50% of women view divorce as morally acceptable? The standard error for the estimate assuming the null hypothesis is true is 0.021.

*Answer:* The observed sample proportion is 0.67 with a standard error of 0.021. If the null is true, then we would expect the sampling distribution of the sample mean to be (approximately) normally distributed with a center of 0.50 and SE of 0.021. The standardized score for the sample proportion is then \[
z = \dfrac{\textrm{statistic} - \textrm{null parameter}}{SE} = \dfrac{0.67 - 0.50}{0.021} = 8.10
\] The observed proportion is 8.1 SEs above the hypothesized value of 0.5.

Note that the randomization distribution should look roughly like this (with the observed proportion denoted with a red X):

```
library(ggplot2)
# Create a data frame with a sequence of x values
x_values <- data.frame(x = seq(0.3, 0.7, length.out = 100))
# Use ggplot2 to plot the normal distribution curve and add the red point
ggplot(x_values, aes(x = x)) +
stat_function(fun = dnorm, args = list(mean = 0.5, sd = 0.021), color = "blue") +
geom_point(aes(x = 0.67, y = 0), color = "red", shape = "X") +
xlab("sample proportions") +
ylab("density") +
theme_minimal()
```

*Answer:* As we can see in the normal plot above, the p-value will be very small because the alternative is looking for big sample proportions. The p-value is the proportion of times we get a sample proportion as big, or bigger than, 0.67; or equivantly, the proportion of times we get a sample proportion that is at least 8.1 SEs above the hypothesized proportion. We would report a p-value that is less than 0.0001.

*Answer:* Without knowing the bootstrap SE, our best guess at it would be from the randomization distribution SE which is given as 0.021. Our 99% confidence interval will look like: \[
statistic \pm z^*SE = 0.67 \pm z^* (0.021)
\] The \(z^*\) for a 99% CI corresponds to the 99.5th percentile (90% in middle + 0.5% in the left tail). With \(z^* = 2.576\), we get a 99% confidence interval of 0.616 to 0.724.

In the same study described above, we find that 71% of men view divorce as morally acceptable. Use this and the information in the previous example to test whether there is a significant difference between men and women in how they view divorce. The standard error for the difference in proportions under the null hypothesis that the proportions are equal is 0.029.

*Answer:* Using the same notation as (1a), except denoting male/female populations, we get

\[ H_0: p_f = p_m \ \ H_A: p_f \neq p_m \]

*Answer:* Suppose we look at the difference \(p_m - p_f\). The observed difference is then 0.04 (0.71 - 0.67). This value is about 1.4 SEs above the hypothesized difference of 0: \[
z = \dfrac{\textrm{statistic} - \textrm{null parameter}}{SE} = \dfrac{(0.71 - 0.67) - 0}{0.029} = 1.379
\]

Note that the randomization distribution for the difference in sample proportions should look roughly like this (with the observed proportion difference denoted with a red X):

```
library(ggplot2)
# Create a data frame with a sequence of x values
x_values <- data.frame(x = seq(-0.1, 0.1, length.out = 100))
# Use ggplot2 to plot the normal distribution curve
ggplot(x_values, aes(x = x)) +
stat_function(fun = dnorm, args = list(mean = 0, sd = 0.029), color = "blue") +
geom_point(aes(x = 0.04, y = 0), color = "red", shape = "X") +
xlab("sample proportions") +
ylab("density") +
theme_minimal()
```

*Answer:* This is a two-tail test. Since the observed difference is less than 2 SEs away from 0 we know that the (two-tailed) p-value should be bigger than 0.05. We see that the p-value is 2(0.084) = 0.168.