Alexa Potter
March 25, 2023
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x dplyr::filter() masks stats::filter()
x dplyr::lag() masks stats::lag()The time between the date a patient was recommended for heart surgery and the surgery date for cardiac patients in Ontario was collected by the Cardiac Care Network (“Wait Times Data Guide,” Ministry of Health and Long-Term Care, Ontario, Canada, 2006). The sample mean and sample standard deviation for wait times (in days) of patients for two cardiac procedures are given in the accompanying table. Assume that the sample is representative of the Ontario population
surgical_procedure sample_size mean_wait_time standard_deviation
1 Bypass 539 19 10
2 Angiography 847 18 9Construct the 90% confidence interval to estimate the actual mean wait time for each of the two procedures. Is the confidence interval narrower for angiography or bypass surgery?
First we need to determine the standard error for procedure.
Then we need to set the confidence interval.
[1] 18.29029 19.70971[1] 17.49078 18.50922The confidence interval for angiography is more narrow compared to the bypass confidence interval.
A survey of 1031 adult Americans was carried out by the National Center for Public Policy. Assume that the sample is representative of adult Americans. Among those surveyed, 567 believed that college education is essential for success. Find the point estimate, p, of the proportion of all adult Americans who believe that a college education is essential for success. Construct and interpret a 95% confidence interval for p.
1-sample proportions test with continuity correction
data: 567 out of 1031, null probability 0.5
X-squared = 10.091, df = 1, p-value = 0.00149
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.5189682 0.5805580
sample estimates:
p
0.5499515 The prop.test shows that 95% of confidence intervals calculated would reflect the true percent of adult Americans that believe college education is essential for success is between 51.89682%-58.05580%.
Suppose that the financial aid office of UMass Amherst seeks to estimate the mean cost of textbooks per semester for students. The estimate will be useful if it is within $5 of the true population mean (i.e. they want the confidence interval to have a length of $10 or less). The financial aid office is pretty sure that the amount spent on books varies widely, with most values between $30 and $200. They think that the population standard deviation is about a quarter of this range (in other words, you can assume they know the population standard deviation). Assuming the significance level to be 5%, what should be the size of the sample
Formula: n = ((z-score * sigma)/ME)^2
z-score for significance level 5% (95% confidence level) = 1.96 sigma (SD)= (200-30)/4 ME = 5
According to a union agreement, the mean income for all senior-level workers in a large service company equals $500 per week. A representative of a women’s group decides to analyze whether the mean income μ for female employees matches this norm. For a random sample of nine female employees, ȳ = $410 and s = 90
Hypothesis: H0: μ = 500 HA: μ ≠ 500
Assumptions: - The data is collected from a representative & randomly selected portion of the total population. - The data is a normal distribution - The two groups have the same population variance (homoskedasticity)
Formula for t-test: t = (X‾ - μ0) / (s / √n) X‾ = 410 μ0 = 500 s = 90 n = 9
p-value:
Since the p-value is less than 0.05 we can reject the null hypothesis at significance level 0.05.
Since the p value is less than 0.05 we can reject the null hypothesis at 0.05 significance level.
(Hint: The P-values for the two possible one-sided tests must sum to 1.)
Since the p value is greater than 0.05 we fail to reject the null hypothesis at 0.05 significance level.
Jones and Smith separately conduct studies to test H0: μ = 500 against Ha: μ ≠ 500, each with n = 1000. Jones gets ȳ = 519.5, with se = 10.0. Smith gets ȳ = 519.7, with se = 10.0.
Jones
t = (ȳ - μ)/se
[1] 1.95[1] 0.05145555Smith
t = (ȳ - μ)/se
The result is statistically significant when the p-value is less than or equal to the alpha level.
At α = 0.05, Jones’ p-value of 0.051 is not statistically significant. At α = 0.05, Smith’s p-value of 0.049 is statistically significant.
Using “P ≤ 0.05” versus “P > 0.05,” can leave a gap in the understanding of the full analysis. While one value is statistically significant, these two are extremely close. It’s important to state at what value you reject or fail to reject the null hypothesis, because both of these would not be statistically significant at α = 0.01.
A school nurse wants to determine whether age is a factor in whether children choose a healthy snack after school. She conducts a survey of 300 middle school students, with the results below. Test at α = 0.05 the claim that the proportion who choose a healthy snack differs by grade level. What is the null hypothesis? Which test should we use? What is the conclusion?
#grade_level <- c('6th', '7th', '8th')
#healthy_snack <- c(31,43,51)
#unhealth_snack <- c(69,57,49)
middleschool_df <- c(31, 43, 51, 69, 57, 49)
middleschool_df <- matrix(middleschool_df, nrow=2, ncol = 3, byrow = TRUE)
colnames(middleschool_df) <- c("6th", "7th", "8th")
rownames(middleschool_df) <- c("health_snack", "unhealth_snack")
print(middleschool_df) 6th 7th 8th
health_snack 31 43 51
unhealth_snack 69 57 49For this analysis we would use the X^2 test as it determines is there is an association between categorical variables. The null hypothesis is that variable 1 is independent of variable 2. In this analysis it would mean that choosing a healthy snack is independent of grade level.
Pearson's Chi-squared test
data: middleschool_df
X-squared = 8.3383, df = 2, p-value = 0.01547With a p-value of 0.015 and alpha level of 0.05, we can reject the null hypothesis. We can conclude at a significance level of 0.05 the association between grade level and choosing a healthy snack are statistically significant.
Per-pupil costs (in thousands of dollars) for cyber charter school tuition for school districts in three areas are shown. Test the claim that there is a difference in means for the three areas, using an appropriate test. What is the null hypothesis? Which test should we use? What is the conclusion?
Area_1 <- c(6.2,9.3,6.8,6.1,6.7,7.5)
Area_2 <- c(7.5,8.2,8.5,8.2,7.0,9.3)
Area_3 <- c(5.8,6.4,5.6,7.1,3.0,3.5)
cyberschool_df <- data.frame(Area_1, Area_2, Area_3)
cyberschool_anova_df <- pivot_longer(cyberschool_df, c(Area_1, Area_2, Area_3), names_to = "Area") %>%
rename("Tuition" = "value")
print(cyberschool_df) Area_1 Area_2 Area_3
1 6.2 7.5 5.8
2 9.3 8.2 6.4
3 6.8 8.5 5.6
4 6.1 8.2 7.1
5 6.7 7.0 3.0
6 7.5 9.3 3.5# A tibble: 18 x 2
Area Tuition
<chr> <dbl>
1 Area_1 6.2
2 Area_2 7.5
3 Area_3 5.8
4 Area_1 9.3
5 Area_2 8.2
6 Area_3 6.4
7 Area_1 6.8
8 Area_2 8.5
9 Area_3 5.6
10 Area_1 6.1
11 Area_2 8.2
12 Area_3 7.1
13 Area_1 6.7
14 Area_2 7
15 Area_3 3
16 Area_1 7.5
17 Area_2 9.3
18 Area_3 3.5The statistical test to use for this data set is an ANOVA as it is comparing the mean of two or more independent groups.
The null hypothesis is that there is no difference in mean across variables
Df Sum Sq Mean Sq F value Pr(>F)
Area 2 25.66 12.832 8.176 0.00397 **
Residuals 15 23.54 1.569
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The P value is 0.00397. This means we can reject the null hypothesis at 0.01. The conclusion we can draw is that there is at least one mean that is different.
---
title: "Homework 2"
author: "Alexa Potter"
description: "DACSS 603"
date: "03/28/2023"
format:
html:
toc: true
code-fold: true
code-copy: true
code-tools: true
categories:
- hw1
- desriptive statistics
- probability
---
```{r}
library(tidyverse)
```
# Question 1
The time between the date a patient was recommended for heart surgery and the surgery date for cardiac patients in Ontario was collected by the Cardiac Care Network (“Wait Times Data Guide,” Ministry of Health and Long-Term Care, Ontario, Canada, 2006). The sample mean and sample standard deviation for wait times (in days) of patients for two cardiac procedures are given in the accompanying table. Assume that the sample is representative of the Ontario population
```{r}
surgical_procedure <- c('Bypass', 'Angiography')
sample_size <- c(539,847)
mean_wait_time <- c(19,18)
standard_deviation <- c(10,9)
heart_surgery_df <- data.frame(surgical_procedure, sample_size, mean_wait_time, standard_deviation)
print(heart_surgery_df)
```
Construct the 90% confidence interval to estimate the actual mean wait time for each of the two procedures. Is the confidence interval narrower for angiography or bypass surgery?
First we need to determine the standard error for procedure.
```{r}
se_bypass <- 10/ sqrt(539)
se_angiography <- 9 / sqrt(847)
```
Then we need to set the confidence interval.
```{r}
confidence_level <- 0.90
```
```{r}
tail_area <- (1-confidence_level)/2
```
```{r}
t_score_bypass <- qt(p = 1-tail_area, df = 539-1)
t_score_angiography <- qt(p = 1-tail_area, df = 847-1)
```
```{r}
CI_bypass <- c(19 - t_score_bypass * se_bypass,
19 + t_score_bypass * se_bypass)
print(CI_bypass)
CI_angiography <- c(18 - t_score_angiography * se_angiography,
18 + t_score_angiography * se_angiography)
print(CI_angiography)
```
The confidence interval for angiography is more narrow compared to the bypass confidence interval.
# Question 2
A survey of 1031 adult Americans was carried out by the National Center for Public Policy. Assume that the sample is representative of adult Americans. Among those surveyed, 567 believed that college education is essential for success. Find the point estimate, p, of the proportion of all adult Americans who believe that a college education is essential for success. Construct and interpret a 95% confidence interval for p.
```{r}
prop.test(567, 1031, conf.level = 0.95)
```
The prop.test shows that 95% of confidence intervals calculated would reflect the true percent of adult Americans that believe college education is essential for success is between 51.89682%-58.05580%.
# Question 3
Suppose that the financial aid office of UMass Amherst seeks to estimate the mean cost of textbooks per semester for students. The estimate will be useful if it is within $5 of the true population mean (i.e. they want the confidence interval to have a length of $10 or less). The financial aid office is pretty sure that the amount spent on books varies widely, with most values between $30 and $200. They think that the population standard deviation is about a quarter of this range (in other words, you can assume they know the population standard deviation).
Assuming the significance level to be 5%, what should be the size of the sample
Formula: n = ((z-score * sigma)/ME)^2
z-score for significance level 5% (95% confidence level) = 1.96
sigma (SD)= (200-30)/4
ME = 5
```{r}
((1.96 * ((200-30)/4))/5)^2
```
# Question 4
According to a union agreement, the mean income for all senior-level workers in a large service company equals $500 per week. A representative of a women’s group decides to analyze whether the mean income μ for female employees matches this norm. For a random sample of nine female employees, ȳ = $410 and s = 90
## A. Test whether the mean income of female employees differs from $500 per week. Include assumptions, hypotheses, test statistic, and P-value. Interpret the result.
Hypothesis:
H0: μ = 500
HA: μ ≠ 500
Assumptions:
- The data is collected from a representative & randomly selected portion of the total population.
- The data is a normal distribution
- The two groups have the same population variance (homoskedasticity)
Formula for t-test:
t = (X‾ - μ0) / (s / √n)
X‾ = 410
μ0 = 500
s = 90
n = 9
```{r}
t_stat <- (410 - 500) / (90 / sqrt(9))
t_stat
```
p-value:
```{r}
2*pt(q=t_stat, df=8)
```
Since the p-value is less than 0.05 we can reject the null hypothesis at significance level 0.05.
## B. Report the P-value for Ha: μ < 500. Interpret.
```{r}
pt(q= t_stat, df=8, lower.tail=TRUE)
```
Since the p value is less than 0.05 we can reject the null hypothesis at 0.05 significance level.
## C. Report and interpret the P-value for Ha: μ > 500.
(Hint: The P-values for the two possible one-sided tests must sum to 1.)
```{r}
pt(q= t_stat, df=8, lower.tail=FALSE)
```
Since the p value is greater than 0.05 we fail to reject the null hypothesis at 0.05 significance level.
# Question 5
Jones and Smith separately conduct studies to test H0: μ = 500 against Ha: μ ≠ 500, each with n = 1000. Jones gets ȳ = 519.5, with se = 10.0. Smith gets ȳ = 519.7, with se = 10.0.
## A. Show that t = 1.95 and P-value = 0.051 for Jones. Show that t = 1.97 and P-value = 0.049 for Smith.
Jones
t = (ȳ - μ)/se
```{r}
jones_t <- (519.5-500)/10
print(jones_t)
#p-value:
2*pt(-abs(jones_t),df=1000-1)
```
Smith
t = (ȳ - μ)/se
```{r}
smith_t <- (519.7-500)/10
print(smith_t)
#p-value:
2*pt(-abs(smith_t),df=1000-1)
```
## B. Using α = 0.05, for each study indicate whether the result is “statistically significant.”
The result is statistically significant when the p-value is less than or equal to the alpha level.
At α = 0.05, Jones' p-value of 0.051 is not statistically significant.
At α = 0.05, Smith's p-value of 0.049 is statistically significant.
## C. Using this example, explain the misleading aspects of reporting the result of a test as “P ≤ 0.05” versus “P > 0.05,” or as “reject H0” versus “Do not reject H0,” without reporting the actual P-value.
Using “P ≤ 0.05” versus “P > 0.05,” can leave a gap in the understanding of the full analysis. While one value is statistically significant, these two are extremely close. It's important to state at what value you reject or fail to reject the null hypothesis, because both of these would not be statistically significant at α = 0.01.
# Question 6
A school nurse wants to determine whether age is a factor in whether children choose a healthy snack after school. She conducts a survey of 300 middle school students, with the results below. Test at α = 0.05 the claim that the proportion who choose a healthy snack differs by grade level. What is the null hypothesis? Which test should we use? What is the conclusion?
```{r}
#grade_level <- c('6th', '7th', '8th')
#healthy_snack <- c(31,43,51)
#unhealth_snack <- c(69,57,49)
middleschool_df <- c(31, 43, 51, 69, 57, 49)
middleschool_df <- matrix(middleschool_df, nrow=2, ncol = 3, byrow = TRUE)
colnames(middleschool_df) <- c("6th", "7th", "8th")
rownames(middleschool_df) <- c("health_snack", "unhealth_snack")
print(middleschool_df)
```
For this analysis we would use the X^2 test as it determines is there is an association between categorical variables. The null hypothesis is that variable 1 is independent of variable 2. In this analysis it would mean that choosing a healthy snack is independent of grade level.
```{r}
chisq.test(middleschool_df, correct = FALSE)
```
With a p-value of 0.015 and alpha level of 0.05, we can reject the null hypothesis. We can conclude at a significance level of 0.05 the association between grade level and choosing a healthy snack are statistically significant.
# Question 7
Per-pupil costs (in thousands of dollars) for cyber charter school tuition for school districts in three areas are shown. Test the claim that there is a difference in means for the three areas, using an appropriate test. What is the null hypothesis? Which test should we use? What is the conclusion?
```{r}
Area_1 <- c(6.2,9.3,6.8,6.1,6.7,7.5)
Area_2 <- c(7.5,8.2,8.5,8.2,7.0,9.3)
Area_3 <- c(5.8,6.4,5.6,7.1,3.0,3.5)
cyberschool_df <- data.frame(Area_1, Area_2, Area_3)
cyberschool_anova_df <- pivot_longer(cyberschool_df, c(Area_1, Area_2, Area_3), names_to = "Area") %>%
rename("Tuition" = "value")
print(cyberschool_df)
print(cyberschool_anova_df)
```
The statistical test to use for this data set is an ANOVA as it is comparing the mean of two or more independent groups.
The null hypothesis is that there is no difference in mean across variables
```{r}
anova_cyberschool <- aov(Tuition ~ Area, data = cyberschool_anova_df)
summary(anova_cyberschool)
```
The P value is 0.00397. This means we can reject the null hypothesis at 0.01. The conclusion we can draw is that there is at least one mean that is different.