Homework 2

The first homework on descriptive statistics and probability

Code

#| label: setup
#| warning: false
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knitr::opts_chunk$set(echo = TRUE, warning=FALSE, message=FALSE)

library(readxl)
library(tidyverse)

── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
✔ ggplot2 3.3.6      ✔ purrr   0.3.4 
✔ tibble  3.1.8      ✔ dplyr   1.0.10
✔ tidyr   1.2.1      ✔ stringr 1.4.1 
✔ readr   2.1.2      ✔ forcats 0.5.2 
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag()    masks stats::lag()

Code

library(ggplot2)
library(dplyr)

1

Creating the table with the given data.

Code

SP <- c('Bypass', 'Angiography')
SS <- c(539, 847)
MW <- c(19, 18)
SD <- c(10, 9)

ServeyData <- data.frame(SP, SS, MW, SD)
ServeyData

           SP  SS MW SD
1      Bypass 539 19 10
2 Angiography 847 18  9

Calculate Standard error

Code

SE <- SD / sqrt(SS)
SE

[1] 0.4307305 0.3092437

calculate the area of the two tails

Code

CL <- 0.90  
#area in each tail of the distribution for 90%
tail_area <- (1-CL)/2
tail_area

[1] 0.05

calculate t-values by using the qt() function

Code

t_score <- qt(p = 1-tail_area, df = SS-1)
t_score

[1] 1.647691 1.646657

calculate the confidence interval

Code

CI <- c(MW - t_score * SE,
        MW + t_score * SE)
print(CI)

[1] 18.29029 17.49078 19.70971 18.50922

The 90% confidence interval for bypass is [18.29, 19.71] days and for angiography it is [17.49, 18.51] days. The confidence interval for angiography is narrower.

2

Using prop.test() to calculate p and the 95% confidence interval.

Code

set.seed(0)
prop.test(567, 1031, conf.level = .95)

    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 95% confidence interval for the point estimate is 0.5195839 - 0.5803191.The point estimate for the proportion of all adult Americans who believe that a college education is essential for success is 0.55.

3

Calculate the min sample size

Code

# calculate population SD.
SD <- (200-30)/4
#margin of error
ME <- (10/2)
# calculate sample size.
samplesize <- ((1.96*SD)/ME)^2
samplesize

[1] 277.5556

the size of the sample should be 278

4

a

calculate t statistic since it will show us the difference in two means

Null hypothesis mean = 500

Code

t_stats <- (410-500)/(90/sqrt(9))
t_stats

[1] -3

Calculate P value

Code

p_value <- 2* pt(t_stats, df=8)
p_value

[1] 0.01707168

The test statistic is -3 and the p-value is 0.01707168.The p-value is substantially less than .05 is the evidence that we can reject the null hypothesis. There is strong evidence that the mean income of female employees is not equal to $500.

b

Code

PL <- pt(t_stats, df = 8, lower.tail = TRUE)
PL

[1] 0.008535841

Since p-value is 0.0085 is less than the alpha level of 0.05, we can reject the null hypothesis. There is evidence that the mean income of female employees is less than $500.

c

Code

PL <- pt(t_stats, df = 8, lower.tail = FALSE)
PL

[1] 0.9914642

Since p-value is 0.991 is more than the alpha level of 0.05, we cannot reject the null hypothesis. There is evidence that the mean income of female employees is more than $500.

5

a

Code

# calculate standard deviation 
Std_Dev <- 10*sqrt(1000)

# calculate t for Jones.
t_jones <- ((519.5-500)/Std_Dev) * sqrt(1000)
t_jones

[1] 1.95

Code

# calculate p-value for Jones.
p_jones <- 2*(pt(q=t_jones, df=999, lower.tail=FALSE))
p_jones

[1] 0.05145555

Code

# calculate t for Smith.
t_smith <- ((519.7-500)/Std_Dev) * sqrt(1000)
t_smith

[1] 1.97

Code

# calculate p-value for Smith.
p_smith <- 2*(pt(q=t_smith, df=999, lower.tail=FALSE))
p_smith

[1] 0.04911426

b

At the .05 significance level, we could say that Jones would be unable to reject the null hypothesis since his exceeds .05. Smith on the other hand would barley be able to reject the null hypothesis with his equalling .049.

c

Both of these p values were extremely close to the actual cut off point which shows including them is important. If I would have saw these p scores I would have had doubts or questions regarding the data and would have ran my own test to validate the claims. I think that is reason it would be important to include them to allow other people to see how close the study was.

6

Code

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)
t.test(gas_taxes, mu = 45, alternative = 'less')

    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 

Here we can see that the p value for this is .038 which means we can reject the null hypothesis that gas prices are equal to or greater than 45 cents. The mean sample that came up was also within the range of the confidence interval.