October 29, 2018
gss2010 <- read.csv('../data/os3_data/Ch 5 Exercise Data/gss2010.csv')
gss2010$degree <- factor(as.character(gss2010$degree),
levels = c('LT HIGH SCHOOL', 'HIGH SCHOOL', 'JUNIOR COLLEGE',
'BACHELOR', 'GRADUATE'),
ordered = TRUE)
tab <- psych::describeBy(gss2010$hrs1, group = gss2010$degree, mat = TRUE)[,c('group1','mean','sd','n')]
tab
## group1 mean sd n ## X11 LT HIGH SCHOOL 38.66942 15.81423 121 ## X12 HIGH SCHOOL 39.59707 14.97125 546 ## X13 JUNIOR COLLEGE 41.39175 18.10361 97 ## X14 BACHELOR 42.54941 13.61731 253 ## X15 GRADUATE 40.84516 15.50540 155
ggplot(gss2010, aes(x = degree, y = hrs1)) + geom_boxplot() +
geom_point(data = tab, aes(x = group1, y = mean), color = 'blue')
a) Write hypotheses for evaluating whether the average number of hours worked varies across the five groups.
\[ H_{0}: \mu_{LTHS} = \mu_{HS} = \mu_{JC} = \mu_{B} = \mu_{G} \]
b) Check conditions and describe any assumptions you must make to proceed with the test.
ggplot(gss2010, aes(x=hrs1, fill=degree,
color=degree)) +
geom_density(alpha = 0.2)
c) Below is part of the output associated with this test. Fill in the empty cells.
summary(aov(hrs1 ~ degree, data = gss2010))
## Df Sum Sq Mean Sq F value Pr(>F) ## degree 4 2006 501.5 2.189 0.0682 . ## Residuals 1167 267382 229.1 ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## 872 observations deleted due to missingness
D) What is the conclusion of the test?
We would fail to reject the null hypothesis.
Two scientists want to know if a certain drug is effective against high blood pressure. The first scientist wants to give the drug to 1,000 people with high blood pressure and see how many of them experience lower blood pressure levels. The second scientist wants to give the drug to 500 people with high blood pressure, and not give the drug to another 500 people with high blood pressure, and see how many in both groups experience lower blood pressure levels. Which is the better way to test this drug?
The GSS asks the same question, below is the distribution of responses from the 2010 survey:
| Response | n |
|---|---|
| All 1000 get the drug | 99 |
| 500 get the drug 500 don’t | 571 |
| Total | 670 |
Parameter of interest: Proportion of all Americans who have good intuition about experimental design.
\[p(population\; proportion)\]
Point estimate: Proportion of sampled Americans who have good intuition about experimental design.
\[\hat{p}(sample\; proportion)\]
What percent of all Americans have good intuition about experimental design, i.e. would answer “500 get the drug 500 don't?"
Using a confidence interval \[point\; estimate \pm ME\]
We know that ME = critical value x standard error of the point estimate. \[SE_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\]
Sample proportions will be nearly normally distributed with mean equal to the population mean, p, and standard error equal to \(\sqrt{\frac{p(1-p)}{n}}\).
\[\hat { p } \sim N\left( mean=p,SE=\sqrt { \frac { p(1-p) }{ n } } \right) \]
This is true given the following conditions:
Given: \(n = 670\), \(\hat{p} = 0.85\).
Conditions:
Independence: The sample is random, and 670 < 10% of all Americans, therefore we can assume that one respondent’s response is independent of another.
Success-failure: 571 people answered correctly (successes) and 99 answered incorrectly (failures), both are greater than 10.
Given: \(n = 670\), \(\hat{p} = 0.85\).
\[0.85 \pm 1.96 \sqrt{\frac{0.85 \times 0.15}{670}} = \left(0.82,\; 0.88\right)\]
We are 95% confidence the true proportion of Americans that have a good intuition about experimental designs is betwee 82% and 88%.
Suppose you want a 3% margin of error, how many people would you have to survey?
Use \(\hat{p} = 0.5\)
\[0.03 = 1.96 \times \sqrt{\frac{0.5 \times 0.5}{n}}\] \[0.03^2 = 1.96^2 \times \frac{0.5 \times 0.5}{n}\] \[n \approx 1,068\]
Scientists predict that global warming may have big effects on the polar regions within the next 100 years. One of the possible effects is that the northern ice cap may completely melt. Would this bother you a great deal, some, a little, or not at all if it actually happened?
| Response | GSS | Duke |
|---|---|---|
| A great deal | 454 | 69 |
| Some | 124 | 40 |
| A little | 52 | 4 |
| Not at all | 50 | 2 |
| Total | 680 | 105 |
Parameter of interest: Difference between the proportions of all Duke students and all Americans who would be bothered a great deal by the northern ice cap completely melting.
\[p_{Duke} - p_{US}\]
Point estimate: Difference between the proportions of sampled Duke students and sampled Americans who would be bothered a great deal by the northern ice cap completely melting.
\[\hat{p}_{Duke} - \hat{p}_{US}\]
Standard error of the difference between two sample proportions
\[SE_{\hat{p}_1 - \hat{p}_2} = \sqrt{ \frac{p_1\left(1 - p_1\right)}{n_1} + \frac{p_2\left(1 - p_2\right)}{n_2} }\]
Conditions:
Construct a 95% confidence interval for the difference between the proportions of Duke students and Americans who would be bothered a great deal by the melting of the northern ice cap (\(p_{Duke} − p_{US}\)).
| Data | Duke | US |
|---|---|---|
| A great deal | 69 | 454 |
| Not a great deal | 36 | 226 |
| Total | 105 | 680 |
| \(\hat{p}\) | 0.657 | 0.668 |
\[ \left(\hat{p}_{Duke} - \hat{p}_{US}\right) \pm z* \times \sqrt{ \frac{p_{Duke}\left(1 - p_{Duke}\right)}{n_{Duke}} + \frac{p_{US}\left(1 - p_{US}\right)}{n_{US}} } \]
\[(0.657 − 0.668) \pm 1.96 \times \sqrt{\frac{0.657 \times 0.343}{105} + \frac{0.668 \times 0.332}{680}} = \left(-0.108,\; 0.086\right)\]