Guanhua Tan
March 23, 2023
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✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()[1] 0.01855288[1] 1.647691[1] 18.96943 19.0305718.97 <= CI_bypass <= 19.03
[1] 0.01062574[1] 1.647691[1] 17.98249 18.0175117.98 <= CI_angiograpy<= 18.02
The Confidence Interval is narrower for Angiography surgery because it has a smaller standard_error.
[1] 0.5499515Error in qt(p - tail_area2, df = 1030): object 'p' not foundError in eval(expr, envir, enclos): object 't_score2' not foundError in eval(expr, envir, enclos): object 't_score2' not foundError in eval(expr, envir, enclos): object 'CI2_A' not foundError in eval(expr, envir, enclos): object 'CI2_B' not found0.549 <= P <= 0.551
the size of students is 277
#Question 4
Null hypothesis: The mean income of female employees is equal to $500 per week. H0: μ = $500 Alternative hypothesis: The mean income of female employees is different from $500 per week. Ha: μ ≠ $500 t.test suggests the mean income of female employees is different from $500 per week. We reject the Null hypothesis.
[1] -3[1] 0.01707168t-statistic is -3. p-value is 0.017.
B. Report the P-value for Ha: μ < 500. Interpret. C. Report and interpret the P-value for Ha: μ > 500.
[1] 0.008535841[1] 0.9914642For Ha: mu<500, we run the pt function and p-value is 0.008, which suggests that we reject the Null hypothesis and the mean income of female employees is much less than 500.
For Ha: mu >500, we run the pt function and p-value is 0.99, which suggests we fail to reject the Null hypothesis and we are unable to demonstrate the income mean of female employees is greater thant 500.
[1] 1.95[1] 0.05145555[1] 1.97[1] 0.04911426B If α=0.5, Smith is statically significant because his p-value is smaller than α. C If we don’t get the actual p-value, we can only conclude that Smith is statically significant without that there is a very tiny difference between two groups. Also, we will ignore that Smith’s p-value is barely smaller than α, which suggests that it is not extremely significant.
Pearson's Chi-squared test
data: df_6[, -1]
X-squared = 8.3383, df = 2, p-value = 0.01547Null hypothesis: means of 3 grades to choose two types of snack are equal. We should use chisq test to test the correlation between grades and the counts of healthy and unhealthy snacks. Chisq suggests that we should reject the null hypothesis because p-value is 0.01547, which is smaller than 0.5. In other words, different grades show differen choices of snacks.
Df Sum Sq Mean Sq F value Pr(>F)
Area 2 25.35 12.674 7.993 0.00433 **
Residuals 15 23.78 1.586
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Null hypothesis: mean of three areas are equal. We should use anova test. Anova test suggests that we should reject the null hypothesis because p-value is 0.0043, which is much smaller than 0.5. In other words, tutions are highly related to areas.
---
title: "Homework 2"
author: "Guanhua Tan"
description: "Homework 2"
date: "03/23/2023"
format:
html:
toc: true
code-fold: true
code-copy: true
code-tools: true
categories:
- hw2
- t.test
---
```{r}
library(tidyverse)
```
# Question 1
```{r, echo=T}
# Bypass
s_mean<-19
s_size <-539
standard_error <- 10/539
standard_error
# t-value
confidence_level <- 0.9
tail_area <- (1-confidence_level)/2
t_score <- qt(p = 1-tail_area, df = s_size-1)
t_score
# plug everything back in
CI <- c(s_mean - t_score * standard_error,
s_mean + t_score * standard_error)
print(CI)
```
18.97 <= CI_bypass <= 19.03
```{r, echo=T}
# Angiography
s_mean_a<-18
s_size_a<-847
standard_error_a <- 9/847
standard_error_a
# t-value
confidence_level <- 0.9
tail_area <- (1-confidence_level)/2
t_score<- qt(p = 1-tail_area, df = s_size-1)
t_score
# plug everything back in
CI_a <- c(s_mean_a- t_score * standard_error_a,
s_mean_a + t_score * standard_error_a)
print(CI_a)
```
17.98 <= CI_angiograpy<= 18.02
The Confidence Interval is narrower for Angiography surgery because it has a smaller standard_error.
# Question 2
```{r, echo=TRUE}
p2 <- 567/1031
p2
SE2 <-sqrt(p2*(1-p2)/1031)
tail_area2 <-(1-0.95)/2
t_score2 <-qt(p-tail_area2, df=1030)
CI2_A<-p2-t_score2*SE2
CI2_B <-p2+t_score2*SE2
CI2_A
CI2_B
```
0.549 <= P <= 0.551
# Question 3
```{r, echo=T}
sd_question3 <- (200-30)/4
Margin3 <-5
n <- (1.96*sd_question3/Margin3)^2
n
```
the size of students is 277
#Question 4
Null hypothesis: The mean income of female employees is equal to $500 per week.
H0: μ = $500
Alternative hypothesis: The mean income of female employees is different from $500 per week.
Ha: μ ≠ $500
t.test suggests the mean income of female employees is different from $500 per week. We reject the Null hypothesis.
```{r}
female_group_mean <-410
sd_4<-90
n_4<-9
t_stat4<-(female_group_mean-500)/(sd_4/sqrt(n_4))
P_value_4 <-(1-pt(t_stat4, df = n_4-1, lower.tail = F))*2
t_stat4
P_value_4
```
t-statistic is -3.
p-value is 0.017.
B. Report the P-value for Ha: μ < 500. Interpret.
C. Report and interpret the P-value for Ha: μ > 500.
```{r}
P_value_lower4<-pt(t_stat4, df=n_4-1, lower.tail=TRUE)
P_value_high4<-pt(t_stat4, df=n_4-1, lower.tail = F)
P_value_lower4
P_value_high4
```
For Ha: mu<500, we run the pt function and p-value is 0.008, which suggests that we reject the Null hypothesis and the mean income of female employees is much less than 500.
For Ha: mu >500, we run the pt function and p-value is 0.99, which suggests we fail to reject the Null hypothesis and we are unable to demonstrate the income mean of female employees is greater thant 500.
# Question 5
```{r}
# Question 5
t_score_5_Jones <-(519.5-500)/10
p_value_5_Jones<- 2*(1-pt(t_score_5_Jones, df=999))
t_score_5_Jones
p_value_5_Jones
t_score_5_Smith <-(519.7-500)/10
p_value_5_Smith <- 2*(1-pt(t_score_5_Smith, df=999))
t_score_5_Smith
p_value_5_Smith
```
B If α=0.5, Smith is statically significant because his p-value is smaller than α.
C If we don't get the actual p-value, we can only conclude that Smith is statically significant without that there is a very tiny difference between two groups. Also, we will ignore that Smith's p-value is barely smaller than α, which suggests that it is not extremely significant.
# Question 6
```{r}
df_6<-data.frame("Grade Level"=c("Heathy sanck", "Unhealth snack"), "6th grade"=c(31,69), "7th grade"=c(43,57), "8th grade"=c(51,49))
chisq.test(df_6[,-1], correct=F)
```
Null hypothesis: means of 3 grades to choose two types of snack are equal.
We should use chisq test to test the correlation between grades and the counts of healthy and unhealthy snacks.
Chisq suggests that we should reject the null hypothesis because p-value is 0.01547, which is smaller than 0.5. In other words, different grades show differen choices of snacks.
# Question 7
```{r}
# Question 7
df_7<- data.frame("Area1"=c(6.2,9.3,6.8,6.1,6.7,7.5),
"Area2"=c(7.5,8.2,8.5,8.2,7.0,9.3),
"Area3"=c(5.8,6.5,5.6,7.1,3.0,3.5))
df_7_long <- df_7 %>%
pivot_longer(cols=c(Area1, Area2, Area3),names_to="Area", values_to = "Fee")
my.anova_7<-aov(Fee ~ Area, df_7_long)
summary(my.anova_7)
```
Null hypothesis: mean of three areas are equal.
We should use anova test.
Anova test suggests that we should reject the null hypothesis because p-value is 0.0043, which is much smaller than 0.5. In other words, tutions are highly related to areas.