HW2

Code

library(tidyverse)
library(readxl)
library(ggplot2)
library(dplyr)
library(tidyr)
knitr::opts_chunk$set(echo = TRUE)

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

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?

Code

bp_num <- 539
bp_mean <- 19
bp_sd <- 10
bp_se <- sqrt(bp_sd)
ag_num <- 847
ag_mean <- 18
ag_sd <- 9
ag_se <- sqrt(ag_sd)

alpha = 0.1
degrees.freedom = bp_num - 1
t.score = qt(p=alpha/2, df=degrees.freedom,lower.tail=F)

margin.error <- t.score * bp_se

lower.bound <- bp_mean - margin.error
upper.bound <- bp_mean + margin.error
bypass_confidenceInterval <- print(c(lower.bound,upper.bound))

[1] 13.78954 24.21046

Code

degrees.freedom = ag_num - 1
t.score = qt(p=alpha/2, df=degrees.freedom,lower.tail=F)

margin.error <- t.score * ag_se

lower.bound <- ag_mean - margin.error
upper.bound <- ag_mean + margin.error
angiography_confidenceInterval <- print(c(lower.bound,upper.bound))

[1] 13.06003 22.93997

The confidence interval is narrower for angiography surgery than it is for bypass surgery. The bypass surgery confidence interval spans from 13.78954 to 24.21046 days while the angiography surgery spans from 13.06003 to 22.93997 days. Since the bypass surgery confidence interval spans 10.42 days and the angiography surgery spans 9.88 days, the angiography surgery has a narrower confidence interval than the bypass surgery.

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.

Code

n = 1031 
college_y = 567
p = college_y/n
print(p)

[1] 0.5499515

Code

margin_error <- qnorm(0.95)*sqrt(p*(1-p)/n)
lowerbound <- p - margin_error
upperbound <- p + margin_error

p_confidenceInterval <- print(c(lowerbound,upperbound))

[1] 0.5244662 0.5754368

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

Code

margin_error_max = 5
confidence_interval_length = 10
lower_bound_avg = 30
upper_bound_avg = 200
range = upper_bound_avg - lower_bound_avg

z = .95 +(1-.95)/2
n = ((z - (range/4))/confidence_interval_length)^2

print(n)

[1] 17.24326

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

Code

income_n <- c(rnorm(9, mean = 410, sd = 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.

assumption: The mean income of female employees is $500 per week. null hypothesis: The mean income of female employees is $500 per week. alternative hypothesis: The mean income of female employees differs from $500 per week.

Code

income_female <- t.test(income_n, mu=500)
print(income_female)

    One Sample t-test

data:  income_n
t = -4.0036, df = 8, p-value = 0.00393
alternative hypothesis: true mean is not equal to 500
95 percent confidence interval:
 303.2482 447.0648
sample estimates:
mean of x 
 375.1565 

The result is that with 95% certainty the mean female income falls between $322.6557 and $441.1327. Since the range falls below $500, we reject the null hypothesis that the mean female income is $500, and accept the alternative hypothesis that the true mean female income is not equal to $500.

B. Report the P-value for Ha: μ < 500. Interpret.

Code

income_female <- t.test(income_n, mu=500, alternative = 'less')
print(income_female)

    One Sample t-test

data:  income_n
t = -4.0036, df = 8, p-value = 0.001965
alternative hypothesis: true mean is less than 500
95 percent confidence interval:
     -Inf 433.1429
sample estimates:
mean of x 
 375.1565 

The p_value is 0.0081. It is less than the 0.5 alpha so we reject the null hypothesis that that the mean female income is $500, and accept the alternative hypothesis that the true mean female income is less than $500.

C. Report and interpret the P-value for Ha: μ > 500.

Code

income_female <- t.test(income_n, mu=500, alternative = 'greater')
print(income_female)

    One Sample t-test

data:  income_n
t = -4.0036, df = 8, p-value = 0.998
alternative hypothesis: true mean is greater than 500
95 percent confidence interval:
 317.1701      Inf
sample estimates:
mean of x 
 375.1565 

The p_value is 0.9919. It is not less than the 0.5 alpha so we fail to reject the null hypothesis that that the mean female income is $500.

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.

Code

sd.js = 10*(sqrt(1000))

income_jones <- c(rnorm(1000, mean = 519.5, sd = sd.js))
incomes_smith <- c(rnorm(1000, mean = 519.7, sd = sd.js))

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.

Code

jones_t <- (519.5 - 500)/(sd.js/sqrt(1000))
jones_p <- 2*(pt(-jones_t, 999))

print(jones_t)

[1] 1.95

Code

print(jones_p)

[1] 0.05145555

Code

smith_t <- (519.7 - 500)/(sd.js/sqrt(1000))
smith_p <- 2*(pt(-smith_t, 999))

print(smith_t)

[1] 1.97

Code

print(smith_p)

[1] 0.04911426

B. Using α = 0.05, for each study indicate whether the result is “statistically significant.”

The Jones study result is not statistically significant because their p-value is above the alpha. The Smith study result is statistically significant because the p-value is below the alpha.

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.

Because in this case the p-value is so close to the alpha, it’s misleading to not report the actual p-value. If someone only read that a null hypothesis was not rejected and not that statistical significance criteria was almost met, they may falsely attribute the strength of confidence of which the null hypothesis was not rejected with. It’s important to report the p-value so that the strength of confidence is known.

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?

x = proportion of students who choose a healthy snack null hypothesis: X is not affected by grade level. alternative hypothesis: X is affected by grade level.

Code

grade <- c(6, 7, 8) 
healthy_prop <- c(31, 43, 51)
unhealthy_prop <- c(69, 57, 49)
student_n <- c(100, 100, 100)

student_data <- data.frame(grade, healthy_prop, unhealthy_prop)

student_t <- prop.test(student_data$healthy_prop, n=student_n)
print(student_t)

    3-sample test for equality of proportions without continuity
    correction

data:  student_data$healthy_prop out of student_n
X-squared = 8.3383, df = 2, p-value = 0.01547
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3 
  0.31   0.43   0.51 

The conclusion is that since the p-value of 0.01547 is below our 0.05 value we reject the null hypothesis that the proportion of students who choose a healthy snack is not affected by grade level in favor of the alternative hypothesis that the proportion of students who choose a healthy snack is affected by grade level.

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?

null hypothesis: There is not a difference in mean per-pupil cost (in thousands of dollars) for cyber charter school tuition for school districts for the three areas. alternative hypothesis: There is a difference in mean per-pupil cost (in thousands of dollars) for cyber charter school tuition for school districts for the three areas.

Code

values <- c(6.2,9.3,6.8,6.1,6.7,7.5,
           7.5,8.2,8.5,8.2,7.0,9.3,
           5.8,6.4,5.6,7.1,3.0,3.5)
area <- c(1,1,1,1,1,1,
          2,2,2,2,2,2,
          3,3,3,3,3,3)

cost_data <- data.frame(values, area)

anova <- aov(values~area, data = cost_data)
summary(anova)

            Df Sum Sq Mean Sq F value Pr(>F)  
area         1  10.45  10.453   4.316 0.0542 .
Residuals   16  38.75   2.422                 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The conclusion is that since the p-value of 0.0542 is above our 0.05 value we fail reject the null hypothesis that there is not a difference in mean per-pupil cost (in thousands of dollars) for cyber charter school tuition for school districts for the three areas in favor of the alternative hypothesis that there is a difference in mean per-pupil cost (in thousands of dollars) for cyber charter school tuition for school districts for the three areas.