Emily Duryea
October 17, 2022
Uploading packages to be used for this assignment:
── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
✔ ggplot2 3.3.6 ✔ purrr 0.3.5
✔ tibble 3.1.8 ✔ dplyr 1.0.10
✔ tidyr 1.2.1 ✔ stringr 1.4.1
✔ readr 2.1.3 ✔ forcats 0.5.2
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()Below is the code used for setting up the degrees of freedom (df = sample size - 1). Thus, Bypass df would be 539 - 1 = 538, and, Angiography would be 847 - 1 = 846.
Below is the code for setting up the mean and standard deviation.
The code for calculating t-score with 90% confidence interval is below.
The code for calculating the standard error (sd/sqrt(sample size)) is below.
The code for calculating the margin of error (T-score multiplied by the standard error) is below.
Below is the code for calculating the upper and lower ranges (add mean to margin of error for upper, subtract for lower).
[1] 18.44732 19.55268[1] 17.60338 18.39662The 90% confidence interval for Bypass is [18.45, 19.55], and for Angio it is [17.60, 18.40]. Thus, Angio has the narrower confidence interval, which is logical to conclude because it has a larger sample size (847 > 539), which reduces the margin of error. Additionally, the standard deviation is smaller (9<10), which signifies less variance.
1-sample proportions test without continuity correction
data: N out of S, null probability PE
X-squared = 6.9637e-30, df = 1, p-value = 1
alternative hypothesis: true p is not equal to 0.5499515
95 percent confidence interval:
0.5194543 0.5800778
sample estimates:
p
0.5499515 The 95% confidence interval is [0.519, 0.580], with a p-value of 0.550.
[1] 277.5454Based on these calculations, the needed sample size would be 278 students for a significance level of 5%.
[1] 30[1] -3[1] 0.01707168It is possible to reject the null hypothesis, as the p-value is statistically significant (0.017), less than 0.05.
[1] 0.008535841The p-value for the lower tail is 0.00854.
[1] 0.9914642The p-value for the upper tail is 0.991.
The total of both tails is equal to 1.
[1] 1.95[1] 1.97[1] 0.05145555[1] 0.04911426If the significance level is 0.05, then Smith has statistically significant study findings, while Jones does not.
Both of the studies could be statistically significant, depending on the significance level. For example, a 0.1 significance level would mean Jones also had statistically significant results. In this case, since the t-scores were so similar (0.2 off), it would not be unreasonable for both studies to reject the null hypothesis.
[1] 40.86278
One Sample t-test
data: gas_taxes
t = -1.8857, df = 17, p-value = 0.03827
alternative hypothesis: true mean is less than 45
95 percent confidence interval:
-Inf 44.67946
sample estimates:
mean of x
40.86278 Because the p-value is 0.03827 on a 95% confidence interval, on the 0.05 significance level, it is possible the null hypothesis that gas prices are equal to or greater than $0.45.
---
title: "Homework 2 - Emily Duryea"
author: "Emily Duryea"
description: "The second homework assignment for DACSS 603"
date: "10/17/2022"
format:
html:
toc: true
code-fold: true
code-copy: true
code-tools: true
categories:
- hw2
- Emily Duryea
---
# Homework 2
Uploading packages to be used for this assignment:
```{r}
library(readxl)
library(tidyverse)
library(dplyr)
```
## Question 1
Below is the code used for setting up the degrees of freedom (df = sample size - 1). Thus, Bypass df would be 539 - 1 = 538, and, Angiography would be 847 - 1 = 846.
```{r}
Bypass_df = 538
Angio_df = 846
```
Below is the code for setting up the mean and standard deviation.
```{r}
Bypass_mean = 19
Bypass_sd = 10
Angio_mean = 18
Angio_sd = 9
```
The code for calculating t-score with 90% confidence interval is below.
```{r}
Bypass_tscore <- qt(p = 0.9, df = Bypass_df)
Angio_tscore <- qt(p = 0.9, df = Angio_df)
```
The code for calculating the standard error (sd/sqrt(sample size)) is below.
```{r}
Bypass_se <- Bypass_sd/sqrt(539)
Angio_se <- Angio_sd/sqrt(847)
```
The code for calculating the margin of error (T-score multiplied by the standard error) is below.
```{r}
Bypass_me <- Bypass_tscore*Bypass_se
Angio_me <- Angio_tscore*Angio_se
```
Below is the code for calculating the upper and lower ranges (add mean to margin of error for upper, subtract for lower).
```{r}
Bypass_low <- Bypass_mean - Bypass_me
Bypass_up <- Bypass_mean + Bypass_me
Angio_low <- Angio_mean - Angio_me
Angio_up <- Angio_mean + Angio_me
Bypass <- c(Bypass_low, Bypass_up)
Angio <- c(Angio_low, Angio_up)
Bypass
Angio
```
The 90% confidence interval for Bypass is \[18.45, 19.55\], and for Angio it is \[17.60, 18.40\]. Thus, Angio has the narrower confidence interval, which is logical to conclude because it has a larger sample size (847 \> 539), which reduces the margin of error. Additionally, the standard deviation is smaller (9\<10), which signifies less variance.
## Question 2
```{r}
# Number who believed college ed is essential for success
N <- 567
# Total sample size
S <- 1031
# Calculating point estimate
PE <- N/S
# Using the function prop.test() to find the confidence interval range & p-value
prop.test(N, S, PE)
```
The 95% confidence interval is \[0.519, 0.580\], with a p-value of 0.550.
## Question 3
```{r}
# Calculating confidence interval of 95%
CI95 <- qnorm(0.025, lower.tail = F)
# Calculating sample size needed using confidence interval equation
Student_sample <- ((170*0.25/5)*CI95)^2
Student_sample
```
Based on these calculations, the needed sample size would be 278 students for a significance level of 5%.
## Question 4
### Part A
```{r}
# Calculating the standard error (sd = 90 sample = 9)
company_se <- 90/sqrt(9)
company_se
# Calculating the t-score
company_tscore <- (410-500)/company_se
company_tscore
# Calculating the p-value (df = 9-1 = 8)
company_pvalue <- (pt(q=-3, df=8))*2
company_pvalue
```
It is possible to reject the null hypothesis, as the p-value is statistically significant (0.017), less than 0.05.
### Part B
```{r}
# Calculating the probability of a random sample with a mean of 410 or less
less_company <- pt(-3, 8)
less_company
```
The p-value for the lower tail is 0.00854.
### Part B
```{r}
# Calculating the probability of a random sample with a mean of 410 or more
more_company <- pt(-3, 8, lower.tail = F)
more_company
```
The p-value for the upper tail is 0.991.
```{r}
total_company <- less_company + more_company
total_company
```
The total of both tails is equal to 1.
## Question 5
### Part A
```{r}
# Calculating t-scores
Jones_tscore <- (519.5-500)/10
Jones_tscore
Smith_tscore <- (519.7-500)/10
Smith_tscore
# Calculating p-values
Jones_pvalue <- (pt(q=1.95, df=999, lower.tail=FALSE))*2
Jones_pvalue
Smith_pvalue <- (pt(q=1.97, df=999, lower.tail=FALSE))*2
Smith_pvalue
```
### Part B
If the significance level is 0.05, then Smith has statistically significant study findings, while Jones does not.
### Part C
Both of the studies could be statistically significant, depending on the significance level. For example, a 0.1 significance level would mean Jones also had statistically significant results. In this case, since the t-scores were so similar (0.2 off), it would not be unreasonable for both studies to reject the null hypothesis.
## Question 6
```{r}
gas_taxes <- c(51.27, 47.43, 38.89, 41.95, 28.61, 41.29, 52.19, 49.48, 35.02, 48.13, 39.28, 54.41, 41.66, 30.28, 18.49, 38.72, 33.41, 45.02)
# Getting the mean
mean_gtaxes <- mean(gas_taxes)
mean_gtaxes
# Conducting t-test
t.test(gas_taxes, mu=45.0, alternative="less")
```
Because the p-value is 0.03827 on a 95% confidence interval, on the 0.05 significance level, it is possible the null hypothesis that gas prices are equal to or greater than \$0.45.