hw2
confidence_interval
hypothesis_testing
emma_narkewicz
Emma Narkewicz HW 2: Confidence Intervals and Hypothesis Testing
Author

Emma Narkewicz

Published

March 28, 2023

Code 
library(tidyverse)
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✔ tibble  3.1.8      ✔ dplyr   1.0.10
✔ tidyr   1.2.0      ✔ stringr 1.4.1 
✔ readr   2.1.2      ✔ forcats 0.5.2 
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
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Question 1

I constructed the 90% confidence interval of each of the two heart procedures manually using the equation: CI = (X bar) ± (t × s/sqrt(n)) and the mean wait time, sample standard deviation, and sample size provided for the Bypass and Angiography heart procedures.

For both procedures I calculated the t-score based on 2 tails, as there was no information indicating we should only focus on the upper or lower tail.

Code 
# Calculating 90% CI  manually of Bypass heart surgery
## sample mean wait time
s_mean_bypass <- 19
## sample sd
s_sd_bypass <- 10
## sample size (n)
s_size_bypass <- 539
## standard error  (sd/swrt(n))
s_se_bypass <- s_sd_bypass / sqrt (s_size_bypass)
s_se_bypass
[1] 0.4307305
Code 
## Specify confidence level 
confidence_level_bypass <- 0.9
##Calculate tail area of 2-tail t-test
tail_area_bypass <- (1 - confidence_level_bypass)/2
tail_area_bypass
[1] 0.05
Code 
#calculation t-score using qt
t_score_bypass <-  qt(p = 1 - tail_area_bypass, df = s_size_bypass - 1)
t_score_bypass
[1] 1.647691
Code 
##Calculating upper & lower CI bypass
CI_bypass <- c(s_mean_bypass - t_score_bypass * s_se_bypass, s_mean_bypass + t_score_bypass * s_se_bypass)
print(CI_bypass)
[1] 18.29029 19.70971
  • The 90% confidence interval for the Bypass wait time is 18.29 days - 19.71 days
Code 
#Calculating the 90% CI manually of angiography heart surgery
##sample mean wait time
s_mean_angio <- 18
##sample sd
s_sd_angio <- 9
## sample size (n)
s_size_angio <- 847
## standard error  (sd/swrt(n)
s_se_angio <- s_sd_angio / sqrt (s_size_angio)
s_se_angio
[1] 0.3092437
Code 
## Specify confidence level 
confidence_level_angio <- 0.9
##Calculate tail area of 2-tail t-test
tail_area_angio <- (1 - confidence_level_angio)/2
tail_area_angio
[1] 0.05
Code 
#calculation t-score using qt
t_score_angio <-  qt(p = 1 - tail_area_angio, df = s_size_angio - 1)
t_score_angio
[1] 1.646657
Code 
##Calculating upper & lower CI bypass
CI_angio <- c(s_mean_angio - t_score_angio * s_se_angio, s_mean_angio + t_score_angio * s_se_angio)
print(CI_angio)
[1] 17.49078 18.50922
  • The 90% confidence interval for the Angiography wait time is 17.49 days - 18.51 days
Code 
#narrower CI
delta_CI_bypass = 19.71 - 18.29
delta_CI_bypass
[1] 1.42
Code 
delta_CI_angio = 18.51 - 17.49
delta_CI_angio
[1] 1.02
  • The confidence interval is narrower for the Angiography (CI delta of 1.02 days) than the confidence interval for the Bypass (CI delta of 1.42 days)

Question 2

To calculate & interpret a 95% CI for the proportion of all adult Americans who believe that a college education is essential for success I used prop.test() because this involves estimating a proportion, not a probability. In this test, a “success” is defined as an adult American reporting a college education is essential for success, with the alternative being a response of an adult American reporting a college education is not essential for success.

The point estimate for the proportion of adult Americans who believe that college is essential is 0.55.

Code 
#Calculate point estimate college is essential

sample_size_survey <- 1031
point_estimate_college_essential <- 567/sample_size_survey
point_estimate_college_essential
[1] 0.5499515

Specific information plugged in was:

  • sample proportion p = point estimate (0.5499515),
  • number sample “successes” x = 567
  • sample size n = 1031
Code 
#Performing prop.test
prop.test(x = 567, n = 1031, p = point_estimate_college_essential)

    1-sample proportions test without continuity correction

data:  567 out of 1031, null probability point_estimate_college_essential
X-squared = 6.9637e-30, df = 1, p-value = 1
alternative hypothesis: true p is not equal to 0.5499515
95 percent confidence interval:
 0.5194543 0.5800778
sample estimates:
        p 
0.5499515

The 95% confidence interval is [0.52, 0.58]

This should be interpreted as 95% of confidence intervals calculated with this procedure would contain the true proportion of adult Americans who believe that a college education 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?

Because we are assuming we know the population sd, this signals we should a 2-sided z-test to calculate the sample size. I can solve for sample size using the z-score through the Margin of Error (MOE) equation that:

MOE = zscore @ confidence level * (sqrt(sd^2/n))

rearranging the equation to solve for n by squaring both sides, multiplying by n, and then dividing by MOE^2 results in the equation:

n = (z^2 * sd^2)/ (MOE^2)

  • MOE = +/- 5
  • pop sd = (200-30)/4 = 42.5
  • at an alpha level 0.05, 2-sided z-score = 1.959964
Code 
#solving for sample size

n = ((1.959964^2) * (42.5^2)) / (5^2) 
n
[1] 277.5454

The ideal sample size 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.

Assumptions of a 1-sample t-test:

  • population is normally distributed

  • observations in our sample are generated independently of one another

Information we are provided is that:

  • Population mean (μ) = 500

  • Sample mean (y bar) = 410

  • Sample sd = 90 s

  • Sample n = 9

Null Hypothesis: (H0) female mean income μ = 500 Alternative Hypothesis: (Ha) female mean income μ ≠ 500

Calculate the t statistic using the equation:

t = (y bar - μ ) / (sd estimate / sqrt(N))

Code 
## Calculating tscore

t =  (410 - 500) / (90/sqrt(9))
t
[1] -3

The t-score is -3

To solve for the p-value from the t-score I will use the pt() function, with a q = the t-statistic of -3, df = 8 (9-1).

It it is important to note that an assumption of the pt() function is you are looking for the probability of the lower tail, not both tails. Because the t score is negative looking at the lower tail is appropriate, but to get the probability for 2-tailed test we need to multiply the lower tail probability by 2.

Code 
#Calculating p-value

p_lower_tail = pt(q = t, df =8, lower.tail = TRUE)
p_lower_tail
[1] 0.008535841

This gives us the probability that μ is less than 500, but that is only 1 tail, so we multiply this problity by 2 to get the p value for our 2-sided t-test

Code 
#Two-side t
p_twotail =  p_lower_tail * 2
p_twotail
[1] 0.01707168

With a 2-tail p value of 0.017, we can confidently reject the null hypothesis that female salary μ = 500 as p< 0.05. This suggests that the true mean salary of female workers is not equal to $500.

B

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

Code 
#P-value Ha: μ < 500

p_lower_tail = pt(q = t, df =8, lower.tail = TRUE)

p_lower_tail
[1] 0.008535841

With a small p-value of 0.00854 we reject the null hypothesis that μ = 500, suggesting that the true mean salary of female workers is less than $500.

C

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

Code 
#P-value Ha: μ > 500

p_upper_tail = pt(q = t, df =8, lower.tail = FALSE)

p_upper_tail
[1] 0.9914642

With a large p-value of 0.99 we fail to reject the null hypothesis that μ = 500, suggesting that the true means salary of female workers is not greater than $500.

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 Show that t = 1.95 and P-value = 0.051 for Jones. Show that t = 1.97 and P-value = 0.049 for Smith.

Because we are provided with standard error not standard deviation, I used the equation:

t = (ȳ - μ) / se

Code 
#Jones T score

t_Jones = (519.5 - 500) / 10.0
t_Jones
[1] 1.95

For Jones the t score is indeed 1.95

To calculate the p-value, I will use pt(), q = 1.95, df = (1000-1) = 999). Because Jone’s t-score is positive, we look at the upper tail.

Code 
#Calculate 1-tail p-value Jones
p_upper_tail_Jones = pt(q = t_Jones, df = 999, lower.tail = FALSE)
p_upper_tail_Jones
[1] 0.02572777

The p value of the upper tail for Jones is 0.02572777, but because the alternative hypothesis is 2-tailed, it needs to be multiplied by 2

Code 
#Calculating p_value Jones
p_Jones = p_upper_tail_Jones * 2
p_Jones
[1] 0.05145555

This shows the p-value of Jones is indeed 0.51

Code 
# Smith T score 
t_Smith = (519.7 - 500) / 10.0
t_Smith
[1] 1.97

The t-score for Smith is indeed 1.97. The p value is calculated the same way as for Jones, using pt() & the upper tail:

Code 
#Calculate 1-tail p-value Smith
p_upper_tail_Smith = pt(q = t_Smith, df = 999, lower.tail = FALSE)
p_upper_tail_Smith
[1] 0.02455713
Code 
#Smith p=value
p_Smith = p_upper_tail_Smith * 2
p_Smith
[1] 0.04911426

Smith’s p-value does = 0.049

B

Using an alpha level = 0.05 for both

  • Jones’ p-value = 0.51 means a non-statistically significant result, where we fail to reject the null hypothesis that μ = 500
  • Smith’s p=value = 0.49 means a statistically significant results, where we reject the null hypothesis that μ = 500

C

This example showcases how reporting the actual p-value is important, as in this case the p-values are very close between Jones & Smith, which are both right on the edge of significance (0.050 +/- 0.01). A p-value of 0.049 & a p-value of 0.0000000049 are both less than 0.05 & would both lead us to reject the null hypothesis, but the second p-value is highly significant while the first p-value is only barely significant at an alpha level of 0.05. Reporting the p-value itself gives readers & other researchers information to much better understand your statistical findings than just reporting if the p value is > or < 0.05 or if you do or do not reject H0.

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?

  • The null hypothesis is that there is no association between the grade of the student and if they choose a healthy snack (they are independent). In this question independence would result in the proportion of students in each grade who choose a healthy snack is the same: H0: p6th = p7th = p8th

  • Because I want to test if there is an association between 2 categorical variables, snack preference & grade level, I used a chi-squared test.

First I recreated the HW2 snacks data table in R:

Code 
#Recreating nurse data

Health_tab <- matrix(c(31, 43, 51, 69, 57, 49), nrow=2, byrow=TRUE)
colnames(Health_tab) <- c('6th','7th','8th')
rownames(Health_tab) <- c('Healthy','Unhealthy')
Health_tab <- as.table(Health_tab)

Health_tab
          6th 7th 8th
Healthy    31  43  51
Unhealthy  69  57  49

Then a chi-squared was run on the data set:

Code 
#Chi Squared test

chisq.test(Health_tab)

    Pearson's Chi-squared test

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

  • Conclusion : The p-value of 0.015 from the chi-squared test is less than alpha=0.05, meaning we reject the null hypothesis that there is no association between student grade level and preference for healthy snacks.

  • At a significance level of 0.05 and df = 2, the critical value = 5.991. The X-squared = 8.3383 is greater than the critical value, once again supporting rejecting the null hypothesis. This suggests snack preference is not independent of grade level.

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 there is no difference between the mean per-pupil cost for cyber charter tuition in 3 different areas, which can be written out as:

H0: μ1 = μ2 = μ3

  • Because we are testing if there is a difference between the means of 3 or more groups that differ by the categorical variable of area we should use a ANOVA test

  • I entered in the data below manually, and then pivoted longer to get 2 columns, Area and Cost.

Code 
Area_1 <- c( 6.2, 9.3, 6.8, 6.1, 6.7, 7.5)
Area_2 <- c( 7.5, 8.2, 8.5, 8.2, 7.0, 9.3)
Area_3 <- c( 5.8, 6.4, 5.6, 7.1, 3.0, 3.5)

charter_data <- data.frame (Area_1, Area_2, Area_3)
Code 
#Pivoted Longer

pivot_charter <- charter_data %>%
  pivot_longer(c(Area_1, Area_2, Area_3 ), names_to= "Area", values_to="Cost")

pivot_charter
# A tibble: 18 × 2
   Area    Cost
   <chr>  <dbl>
 1 Area_1   6.2
 2 Area_2   7.5
 3 Area_3   5.8
 4 Area_1   9.3
 5 Area_2   8.2
 6 Area_3   6.4
 7 Area_1   6.8
 8 Area_2   8.5
 9 Area_3   5.6
10 Area_1   6.1
11 Area_2   8.2
12 Area_3   7.1
13 Area_1   6.7
14 Area_2   7  
15 Area_3   3  
16 Area_1   7.5
17 Area_2   9.3
18 Area_3   3.5

Lastly, I ran the ANOVA, specifying Area as the independent variable & Cost (per pupil) as the dependent variable.

Code 
#Anova 
anova_charter <- aov( Cost ~ Area, data = pivot_charter)

summary(anova_charter)
            Df Sum Sq Mean Sq F value  Pr(>F)   
Area         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

The resulting p-value of 0.00397 is less than an alpha = 0.05 and therefor significant, leading us to reject out null hypothesis that there is no difference in the mean per-pupil charter school cost between the 3 areas. There does appear to be a difference in per-pupil charter school cost based on which of the 3 areas the school is in.