Homework 2

Homework 2 for DACSS 603

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.

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?

First we’ll calculate the confidence interval manually with the formula: CI = (X bar) ± (t × s/sqrt(n))

We’ll create objects in R to plug into this formula. We’ll start with the Bypass data.

Code

library(tidyverse)

── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
✔ ggplot2 3.4.0      ✔ purrr   1.0.1 
✔ tibble  3.1.8      ✔ dplyr   1.0.10
✔ tidyr   1.3.0      ✔ stringr 1.5.0 
✔ readr   2.1.3      ✔ forcats 0.5.2 
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag()    masks stats::lag()

Code

# Bypass data
BypassSample <- 539
BypassMean <- 19
BypassSD <- 10

# standard error
BypassSEM <- BypassSD/sqrt(BypassSample)

# t-value
ConfidenceLevel <- .90
BYPtail <- (1-ConfidenceLevel)/2
Bypass_t_score <- qt(1-BYPtail, df=BypassSample-1)

# plug everything back into the CI formula:
CIBypass <- c(BypassMean-Bypass_t_score*BypassSEM, 
              BypassMean+Bypass_t_score*BypassSEM)

print(CIBypass)

[1] 18.29029 19.70971

Now let’s do the same for the Angiography data.

Code

AngioSample <- 847
AngioMean <- 18
AngioSD <- 9

# standard error
AngioSEM <- AngioSD/sqrt(AngioSample)

# t-value
ConfidenceLevel <- .90
Angtail <- (1-ConfidenceLevel)/2
Angio_t_score <- qt(1-Angtail, df=AngioSample-1)

# plug everything back in

CIAngio <- c(AngioMean-Angio_t_score*AngioSEM, 
             AngioMean+Angio_t_score*AngioSEM)

print(CIAngio)

[1] 17.49078 18.50922

Now we can compare the confidence intervals.

Code

abs(diff(CIAngio)) 

[1] 1.018436

Code

abs(diff(CIBypass))

[1] 1.419421

The confidence interval for Angiography is narrower than for Bypass.

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.

We can use the prop.test() function to find the point estimate and the confidence interval:

Code

prop.test(x=567, n=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 

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.

We have to solve for n here. Because we have the standard deviation, we’ll use the CI formula that uses the z-score: CI = (X bar) ± (z × s/sqrt(n))

This can be rewritten: n=((s*z)/5)^2

Using the data we have:

Code

Q3_s = (200-30)/4
Q3_z = qnorm(.975)

((Q3_s*Q3_z)/5)^2

[1] 277.5454

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

4.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.

To solve this problem I followed the steps of hypothesis testing:

1. Specify the null hypothesis and alternative hypothesis

2. Set the significance level (alpha)

3. Calculate the test statistic

4. Compare test statistics to the critical value determined by alpha, or check if p-value is smaller than alpha

5. Draw a conclusion

1. Our null hypothesis is that the population mean is 500 dollars, our alternative hypothesis is that the population mean is NOT $500

2. The significance level (alpha) is .05, or 5%

3. Calculate the test statistic:

Code

Q4_SD <- 90
Q4_ymean <- 410
Q4_umean <- 500
Q4_s_size <- 9
Q4_SEM <- Q4_SD/sqrt(Q4_s_size)

Q4_T_Statistic <- (Q4_ymean - Q4_umean)/(Q4_SEM)
Q4_T_Statistic

[1] -3

4. Calculate the critical value so that we can compare it with the test statistic.

Code

Q4_Crit_Value <- qt(0.025, df=(Q4_s_size-1))
Q4_Crit_Value

[1] -2.306004

The critical value (on the lower tail end) is -2.3060. Because the test statistic is smaller than the critical value (and larger in absolute terms), we can conclude that we should reject the null hypothesis.

4. (continued) calculate the p-value to check to see if it is smaller than alpha.

Code

Q4_p_value <- pt(Q4_T_Statistic, df = (Q4_s_size-1), lower.tail = TRUE)*2
Q4_p_value

[1] 0.01707168

The p-value is 0.01707, which is smaller than our alpha (.05). 0.01707 < 0.05 This again means that we can reject the null hypothesis.

5. Draw a conclusion

Because the test statistic (-3) is in absolute terms larger than the critical values (2.30) AND because the p-value (0.01707) is less than the alpha levelof .05, we can reject the null hypothesis

Note that when we calculate the confidence interval of these values we can also see that the estimated population mean here is outside of a 95% CI.

To calculate the confidence level we’ll first create some objects we’ll need in our formulas:

Code

ConfLevel <- 0.95
Q4_tail_area <- (1-ConfLevel)/2
Q4_tscore <- qt(p=1-Q4_tail_area, df = Q4_s_size-1)

# Then, using this and some of the objects we created earlier we'll plug everything in: 
Q4_CI <- c(Q4_ymean - (Q4_tscore * Q4_SEM), Q4_ymean + (Q4_tscore * Q4_SEM))

print(Q4_CI)

[1] 340.8199 479.1801

We can see here that $500 is not within the 95% confidence interval, which is another reason we can reject the null hypothesis.

4.B

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

To calculate a one sided P-value you do not need to multiply the formula by 2 and you need to specify that we’re only looking at one side. An alternative hypothesis of μ < 500 means the null hypothesis is μ >= 500. We need to specify that we are looking at the lower tail.

Code

Q4_p_value1 <- pt(Q4_T_Statistic, df = (Q4_s_size-1), lower.tail = TRUE)
Q4_p_value1

[1] 0.008535841

The p-value here is even lower than it was for the two sided test, it is 0.0085. We can definitely reject the null hypothesis that the true population mean is over $500.

4.C

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

To calculate the P-value for Ha: μ > 500 we run the same formula as before but we specify that we’re not interested in the lower tail.

Code

Q4_p_value2 <- pt(Q4_T_Statistic, df = (Q4_s_size-1), lower.tail = FALSE)
Q4_p_value2

[1] 0.9914642

The p-value for this one sided test is quite large. It means that we should not reject the null hypothesis that the true population mean is less than $500.

To test these p-values we can add them up.

Code

Q4_p_value1 + Q4_p_value2

[1] 1

We find that they add up to 1.

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.

5.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.

First we can create some objects to represent the values for this question. Then we can use those to get the test statistic and the p-value for Jones and Smith.

Code

Q5_SEM <- 10 # they have the same standard error
Jones_ymean <- 519.5
Smith_ymean <- 519.7
Q5_s_size <- 1000 # they have the same sample size

Q5_umean <- 500

# calculate the test statistic and p-value for Jones:
Jones_T_Statistic <- (Jones_ymean - Q5_umean)/(Q5_SEM)
Jones_T_Statistic

[1] 1.95

Code

# we find that Jones' test statistic is 1.95. We can use that to find the p-value:
Jones_p_value <- pt(Jones_T_Statistic, df = (Q5_s_size-1), lower.tail = FALSE)*2
Jones_p_value

[1] 0.05145555

Code

# Calculate test statistic and p-value for Smith:
Smith_T_Statistic <- (Smith_ymean - Q5_umean)/(Q5_SEM)
Smith_T_Statistic

[1] 1.97

Code

# we find that Smith's test statistic is 1.97. We can use that to find the p-value:
Smith_p_value <- pt(Smith_T_Statistic, df = (Q5_s_size-1), lower.tail = FALSE)*2
Smith_p_value

[1] 0.04911426

5.B

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

Above we found the p-value for both Jones and Smith, we can use that to show how those values compare to alpha set at 0.05.

Code

alpha <- 0.05

Jones_p_value < alpha

[1] FALSE

Code

Smith_p_value < alpha

[1] TRUE

This shows that technically Jones’ p-value is not statistically significant because it is more than the alpha of 0.05, but Smith’s p-value is technically statistically significant because it is less than the alpha.

5.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.

This question demonstrates how close two results can be and still technically not have the same significance level. In these cases it is important to include the p-value so that readers can put the results into context.

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?

Since we’re dealing with categorical variables (grade and the choice of a healthy or unhealthy snack) we should use a chi-square test. First we’ll create the dataframe we’ll need to run the test in r.

Code

obs <- matrix(c(31, 69, 43, 57, 51, 49), nrow=3, byrow = TRUE)
dimnames(obs) <- list(Grade = c("Grade_6", "Grade_7", "Grade_8"),
                      Snack = c("healthy", "unhealthy"))

obs

         Snack
Grade     healthy unhealthy
  Grade_6      31        69
  Grade_7      43        57
  Grade_8      51        49

Then we can run the chi-square test and print the result:

Code

ChiResult <- chisq.test(obs)

print(ChiResult)

    Pearson's Chi-squared test

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

Given the p-value < 0.05, we can reject the null hypothesis that grade level does not have an effect on the choice of healthy snack. The relatively high X-squared value also indicates that there is a meaningful difference in the choices made by kids in different grade levels.

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?

The null hypothesis is that all three districts have equal average tuition. We can use the ANOVA test since we are comparing three means.

First we’ll create the dataframe in r:

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)

Q7df <- data.frame(Area1, Area2, Area3)

Q7df_PL <- pivot_longer(Q7df, cols = c(Area1, Area2, Area3), names_to = "district", values_to = "tuition")
Q7df_PL

# A tibble: 18 × 2
   district tuition
   <chr>      <dbl>
 1 Area1        6.2
 2 Area2        7.5
 3 Area3        5.8
 4 Area1        9.3
 5 Area2        8.2
 6 Area3        6.4
 7 Area1        6.8
 8 Area2        8.5
 9 Area3        5.6
10 Area1        6.1
11 Area2        8.2
12 Area3        7.1
13 Area1        6.7
14 Area2        7  
15 Area3        3  
16 Area1        7.5
17 Area2        9.3
18 Area3        3.5

Now that we have a dataframe with one continuous variable and one categorical variable, we can use the anova test function in r aov() to test our hypothesis. This type of hypothesis test is appropriate for this scenario because we want to compare the mean of three groups.

Code

Q7o <- aov(tuition ~ district, data = Q7df_PL)
summary(Q7o)

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

Given that the p value here is less than 0.01, we can reject the null hypothesis that the mean tuition is the same at all three schools.