Homework 2

hw2

confidence intervals

hypothesis testing

chi-square

anova

Confidence intervals and hypothesis testing homework for DACSS 603.

Author

Miguel Curiel

Published

March 28, 2023

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.

Surgical Procedure	Sample Size	Mean Wait Time	Standard Deviation
Bypass	539	19	10
Angiography	847	18	9

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

# setting seed for reproducible results
set.seed(0)

# 90% confidence interval for bypass
bypass_ci <- t.test(x=c(rnorm(539, mean=19, sd=10)), conf.level=0.90)$conf.int

# 90% confidence interval for bypass
angiography_ci <- t.test(x=c(rnorm(847, mean=18, sd=9)), conf.level=0.90)$conf.int

# print results
cat("Bypass 90% confidence interval:", bypass_ci, "\n"
    , "Angiography 90% confidence interval:", angiography_ci, "\n")

Bypass 90% confidence interval: 18.32586 19.74194 
 Angiography 90% confidence interval: 17.09532 18.10881

As we can see from the results, bypass’ 90% confidence interval is 18.6-20.1, while angiography’s is 17.5-18.5. Therefore, angiography’s confidence interval is narrower.

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.

Code

# setting seedd for reproducible results
set.seed(0)

# point estimate
point_estimate <- 567/1031

# 95% confidence interval
proportion_test <- prop.test(567, 1031, conf.level = 0.95)$conf.int

# print results
cat("Point estimate:", point_estimate, "\n"
    , "95% confidence interval:", proportion_test, "\n")

Point estimate: 0.5499515 
 95% confidence interval: 0.5189682 0.580558

The results indicate that we are 95% confident that the true proportion of all adult Americans who believe that college education is essential for success lies between .52-.58. In other words, we can say with 95% certainty that 52%-58% of adult Americans believe that college is essential for success.

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?

Code

# margin of error formula
## ME = z * (sigma / sqrt(n))
### solving for n
#### n = (z * sigma/ME)^2
##### replacing values
###### z = significance level 5% = critical z value for 95% conf int = 1.96
###### sigma = 1/4 of range of textbook costs = (200-30)/4 = 42.5
###### ME = estimate within 5 of the true population mean = 5

# calculate result
n <- round((1.96*42.5/5)^2)

# print result
cat("Ideal sample size:", n, "\n")

Ideal sample size: 278

Using the margin of error formula for confidence intervals and solving for n (sample size), we see that the ideal sample size (rounded to the nearest integer) is 278.

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.

Test whether the mean income of female employees differs from $500 per week. Include assumptions, hypotheses, test statistic, and P-value. Interpret the result.
1. Assumptions:
  1. The data is normally distributed.
  2. The sample is a simple random sample.
  3. The standard deviation of the population is unknown.
2. Hypotheses:
  1. H0: μ = 500
  2. H1: Ha: μ ≠ 500
3. Test statistic:
  1. t = (ȳ - μ) / (s / sqrt(n)) = (410 - 500) / (90 / sqrt(9)) = -3
4. P-value:
  1. The P-value is 0.01707168, which is less than the level of significance of 0.05. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees differs from $500 per week.
Report the P-value for Ha: μ < 500. Interpret.
1. The P-value is 0.008535841, which is less than the level of significance of 0.05. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees is less than $500 per week.
Report and interpret the P-value for Ha: μ > 500.
1. The P-value is 0.9914642, which is more than the level of significance of 0.05. Therefore, we fail to reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees is greater than $500 per week.

Code

# A.3.a. Test statistic
(410 - 500) / (90 / sqrt(9))

# A.4.a. P-value two-sided
pt(-3, 8) * 2

# B.1. P-value μ < 500
pt(-3, 8)

# C.1. P-value μ > 500
pt(-3, 8, lower.tail=FALSE)

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.

Show that t = 1.95 and P-value = 0.051 for Jones. Show that t = 1.97 and P-value = 0.049 for Smith.
1. Jones:
  1. t = (ȳ - μ) / se = (519.5 - 500) / 10 = 19.5 / 10 = 1.95
  2. P-value = round(2 * pt(q=1.95, df=999, lower.tail=FALSE), 3) = 0.051
2. Smith:
  1. t = (ȳ - μ) / se = (519.7 - 500) / 10 = 19.7 / 10 = 1.97
  2. P-value = round(2 * pt(q=1.97, df=999, lower.tail=FALSE), 3) = 0.049
Using α = 0.05, for each study indicate whether the result is “statistically significant.”
1. Since Jones’ P-value is greater than 0.05, we fail to reject the null hypothesis, indicating that his results are not statistically significant. In contrast, Smith’s P-value is less than 0.05, therefore we reject the null hypothesis and find his results statistically significant.
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.
1. “P ≤ 0.05” or “reject H0” without reporting the actual P-value can be misleading because it doesn’t provide information on how strong the evidence is against the null hypothesis. For example, both Jones and Smith barely pass the 0.05 threshold, having 0.049 and 0.051, respectively. Reporting this would help readers and analysts to take the strength of the evidence in consideration (in this case, “rejecting H0” or “failing to reject H0” should be taken with a grain of salt).

Code

# A.1.b. Proving Jones' p-value
round(2 * pt(q=1.95, df=999, lower.tail=FALSE), 3)

# A.2.b. Proving Smith's p-value
round(2 * pt(q=1.97, df=999, lower.tail=FALSE), 3)

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?

Grade level	6th grade	7th grade	8th grade
Healthy snack	31	43	51
Unhealthy snack	69	57	49

Null hypothesis: There is no difference in the proportion of students who choose a healthy snack based on grade level.
Test: Chi-squared test because we are assessing whether proportions of outcomes (choosing healthy versus unhealthy snacks) in each grade are equal or different.
Conclusion: Since the p-value is 0.01547, we reject the null hypothesis at the 0.05 level of significance and conclude that there is a significant difference in the proportion of healthy snack choices among the different grade levels.

Code

# create a matrix of the observed values
observed <- matrix(c(31, 43, 51, 69, 57, 49), nrow = 2, byrow = TRUE)

# perform the chi-squared test
result <- chisq.test(observed)

# print the results
print(observed)

     [,1] [,2] [,3]
[1,]   31   43   51
[2,]   69   57   49

Code

print(result)


    Pearson's Chi-squared test

data:  observed
X-squared = 8.3383, df = 2, p-value = 0.01547

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?

Area 1	6.2	9.3	6.8	6.1	6.7	7.5
Area 2	7.5	8.2	8.5	8.2	7.0	9.3
Area 3	5.8	6.4	5.6	7.1	3.0	3.5

Null hypothesis: There is no difference in means for the three areas.
Test: Analysis of Variance (ANOVA) because we are computing the difference between the means of three or more groups.
Conclusion: Given that the P-value associated to the F-statistic is 0.00397, we reject the null hypothesis and conclude that there is a significant difference in means for the three areas.

Code

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, 6.4, 5.6, 7.1, 3.0, 3.5)

anova_result <- aov(c(area1, area2, area3) ~ rep(c("Area 1", "Area 2", "Area 3")
                                                 , c(6, 6, 6)))
print(summary(anova_result))

                                                 Df Sum Sq Mean Sq F value
rep(c("Area 1", "Area 2", "Area 3"), c(6, 6, 6))  2  25.66  12.832   8.176
Residuals                                        15  23.54   1.569        
                                                  Pr(>F)   
rep(c("Area 1", "Area 2", "Area 3"), c(6, 6, 6)) 0.00397 **
Residuals                                                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

--- title: "Homework 2" author: "Miguel Curiel" description: "Confidence intervals and hypothesis testing homework for DACSS 603." date: "03/28/2023" format: html: toc: true code-fold: true code-copy: true code-tools: true categories: - hw2 - confidence intervals - hypothesis testing - chi-square - anova editor: markdown: wrap: 72 --- # 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. | Surgical Procedure | Sample Size | Mean Wait Time | Standard Deviation | |--------------------|-------------|----------------|--------------------| | Bypass | 539 | 19 | 10 | | Angiography | 847 | 18 | 9 | 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? ```{r, eval=TRUE, echo=TRUE} # setting seed for reproducible results set.seed(0) # 90% confidence interval for bypass bypass_ci <- t.test(x=c(rnorm(539, mean=19, sd=10)), conf.level=0.90)$conf.int # 90% confidence interval for bypass angiography_ci <- t.test(x=c(rnorm(847, mean=18, sd=9)), conf.level=0.90)$conf.int # print results cat("Bypass 90% confidence interval:", bypass_ci, "\n" , "Angiography 90% confidence interval:", angiography_ci, "\n") ``` As we can see from the results, bypass' 90% confidence interval is 18.6-20.1, while angiography's is 17.5-18.5. Therefore, angiography's confidence interval is narrower. # 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, eval=TRUE, echo=TRUE} # setting seedd for reproducible results set.seed(0) # point estimate point_estimate <- 567/1031 # 95% confidence interval proportion_test <- prop.test(567, 1031, conf.level = 0.95)$conf.int # print results cat("Point estimate:", point_estimate, "\n" , "95% confidence interval:", proportion_test, "\n") ``` The results indicate that we are 95% confident that the true proportion of all adult Americans who believe that college education is essential for success lies between .52-.58. In other words, we can say with 95% certainty that 52%-58% of adult Americans believe that college is essential for success. # 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? ```{r, eval=TRUE, echo=TRUE} # margin of error formula ## ME = z * (sigma / sqrt(n)) ### solving for n #### n = (z * sigma/ME)^2 ##### replacing values ###### z = significance level 5% = critical z value for 95% conf int = 1.96 ###### sigma = 1/4 of range of textbook costs = (200-30)/4 = 42.5 ###### ME = estimate within 5 of the true population mean = 5 # calculate result n <- round((1.96*42.5/5)^2) # print result cat("Ideal sample size:", n, "\n") ``` Using the margin of error formula for confidence intervals and solving for n (sample size), we see that the ideal sample size (rounded to the nearest integer) is 278. # 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. 1. Assumptions: a. The data is normally distributed. b. The sample is a simple random sample. c. The standard deviation of the population is unknown. 2. Hypotheses: a. H0: μ = 500 b. H1: Ha: μ ≠ 500 3. Test statistic: a. t = (ȳ - μ) / (s / sqrt(n)) = (410 - 500) / (90 / sqrt(9)) = -3 4. P-value: a. The P-value is 0.01707168, which is less than the level of significance of 0.05. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees **differs** from \$500 per week. B. Report the P-value for Ha: μ \< 500. Interpret. 1. The P-value is 0.008535841, which is less than the level of significance of 0.05. Therefore, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees **is less than** \$500 per week. C. Report and interpret the P-value for Ha: μ \> 500. 1. The P-value is 0.9914642, which is more than the level of significance of 0.05. Therefore, we fail to reject the null hypothesis and conclude that there is sufficient evidence to suggest that the mean income of female employees **is greater than** \$500 per week. ```{r, eval=TRUE, echo=TRUE, results='hide'} # A.3.a. Test statistic (410 - 500) / (90 / sqrt(9)) # A.4.a. P-value two-sided pt(-3, 8) * 2 # B.1. P-value μ < 500 pt(-3, 8) # C.1. P-value μ > 500 pt(-3, 8, lower.tail=FALSE) ``` # 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. 1. Jones: a. t = (ȳ - μ) / se = (519.5 - 500) / 10 = 19.5 / 10 = 1.95 b. P-value = round(2 \* pt(q=1.95, df=999, lower.tail=FALSE), 3) = 0.051 2. Smith: a. t = (ȳ - μ) / se = (519.7 - 500) / 10 = 19.7 / 10 = 1.97 b. P-value = round(2 \* pt(q=1.97, df=999, lower.tail=FALSE), 3) = 0.049 B. Using α = 0.05, for each study indicate whether the result is "statistically significant." 1. Since Jones' P-value is greater than 0.05, we fail to reject the null hypothesis, indicating that his results are **not statistically significant**. In contrast, Smith's P-value is less than 0.05, therefore we reject the null hypothesis and his results **are 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. 1. "P ≤ 0.05" or "reject H0" without reporting the actual P-value can be misleading because it doesn't provide information on how strong the evidence is against the null hypothesis. For example, both Jones and Smith barely pass the 0.05 threshold, having 0.049 and 0.051, respectively. Reporting this would help readers and analysts to take the strength of the evidence in consideration (in this case, "rejecting H0" or "failing to reject H0" should be taken with a grain of salt). ```{r, eval=TRUE, echo=TRUE, results='hide'} # A.1.b. Proving Jones' p-value round(2 * pt(q=1.95, df=999, lower.tail=FALSE), 3) # A.2.b. Proving Smith's p-value round(2 * pt(q=1.97, df=999, lower.tail=FALSE), 3) ``` # 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? | Grade level | 6th grade | 7th grade | 8th grade | |-----------------|-----------|-----------|-----------| | Healthy snack | 31 | 43 | 51 | | Unhealthy snack | 69 | 57 | 49 | - Null hypothesis: There is no difference in the proportion of students who choose a healthy snack based on grade level. - Test: Chi-squared test because we are assessing whether proportions of outcomes (choosing healthy versus unhealthy snacks) in each grade are equal or different. - Conclusion: Since the p-value is 0.01547, we reject the null hypothesis at the 0.05 level of significance and conclude that there is a significant difference in the proportion of healthy snack choices among the different grade levels. ```{r, eval=TRUE, echo=TRUE} # create a matrix of the observed values observed <- matrix(c(31, 43, 51, 69, 57, 49), nrow = 2, byrow = TRUE) # perform the chi-squared test result <- chisq.test(observed) # print the results print(observed) print(result) ``` # 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? | | | | | | | | |--------|-----|-----|-----|-----|-----|-----| | Area 1 | 6.2 | 9.3 | 6.8 | 6.1 | 6.7 | 7.5 | | Area 2 | 7.5 | 8.2 | 8.5 | 8.2 | 7.0 | 9.3 | | Area 3 | 5.8 | 6.4 | 5.6 | 7.1 | 3.0 | 3.5 | - Null hypothesis: There is no difference in means for the three areas. - Test: Analysis of Variance (ANOVA) because we are computing the difference between the means of three or more groups. - Conclusion: Given that the P-value associated to the F-statistic is 0.00397, we reject the null hypothesis and conclude that there is a significant difference in means for the three areas. ```{r, eval=TRUE, echo=TRUE} 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, 6.4, 5.6, 7.1, 3.0, 3.5) anova_result <- aov(c(area1, area2, area3) ~ rep(c("Area 1", "Area 2", "Area 3") , c(6, 6, 6))) print(summary(anova_result)) ```