Practice with Normal and t-distribution quantiles

We saw in the previous lab that the pnorm() function could be used to find probabilities associated with specific values for the Normal distribution.

pnorm(-1.96); pnorm(1.96, lower.tail=F)
## [1] 0.0249979
## [1] 0.0249979

i.e.: the probability below -1.96 and above 1.96 are both .025 on a standard normal distribution

The R functions qnorm() and qt() do the opposite: They give the cutoff that corresponds to a probability that we put in the function.

In order to use qnorm() we must give it

  1. the probability
  2. the mean (optional)
  3. the standard deviation (optional)

Note: If you do not specify the mean and standard deviation qnorm() defaults to the standard normal distribution N(0,1).

qnorm(.025)
## [1] -1.959964
qnorm(.975)
## [1] 1.959964

These correspond to the cutoffs we use in a 95% confidence interval when we know \(\sigma\)’s value. These also correspond to the middle 95% of the standard normal distribution because .975-.025 = .95

When we want to use a different confidence % for our CI, we need to use values different from .025 and .975.

We will do this by using \(\alpha\) for a 100(1-\(\alpha\))% CI. The values will be \(\alpha / 2\) and \(1-(\alpha / 2)\).

For the 95% CI we have \(\alpha = .05\) and so that gave us the cutoffs \(\alpha / 2 = .05 / 2 = .025\) and \(1-(\alpha / 2)= 1-(.05 / 2) = .975\). We will actually get the same value just with a different sign (positive or negative) for each of these, so we will only care about the positive one from here on out.

The qt() function will work very similar. We input the cutoffs according the the confidence % with the \(\alpha\) values and it gives us the corresponding quantiles of the t-distribution instead of the normal distribution. We also need to include the degrees of freedom for the normal distribution.

# 95\% CI quantile cutoffs
# for a sample with sample size = 34 (n=34)
qt(.975, df=33)
## [1] 2.034515

This tells us we need to add and subtract 2.03 SE’s in our 95% CI for the mean using the t-distribution with df=33.

Use the qnorm() and qt() functions to answer the following:

Question 1

Part A: How many SE’s will we add and subtract for a 99% CI using the normal distribution?

qnorm(.995)
## [1] 2.575829

Part B: How many SE’s will we add and subtract for a 90% CI using the normal distribution?

qnorm(.95)
## [1] 1.644854

Part C: How many SE’s will we add and subtract for a 95% CI using the t-distribution with a sample size of 30?

qt(.975, df=29)
## [1] 2.04523

Part D: How many SE’s will we add and subtract for a 95% CI using the t-distribution with a sample size of 200?

qt(.975, df=199)
## [1] 1.971957

Part E: Using your answer to A and B, what do we notice about the relationship between confidence level and interval width?

# The larger the confidence level the wider the interval is

Part F: How does the width of CI made using the t-distribution compare to the width of a CI made using the Normal distribution (for the same confidence level)?

# the CI using the t-distribution is a little bit wider than the CI from the normal distribution when the confidence level is the same

Mercury in Fish (revisited) Pt. 1

We are going to work on a problem similar to the yellowfin tuna and mercury levels we covered in the previous lab. Before, we knew the values for the population mean and standard deviation. Most of the time we will not know these.

For this example, suppose the pop. mean and standard deviation are not known. (\(\mu\) = ?, \(\sigma\) = ?). Do not use the values from the previous lab.

Research Question: What is the pop. mean mercury level (micro-grams mercury per gram of fish) of yellowfin tuna?

To answer this question, scientists recorded the mercury levels in 48 randomly selected yellowfin tuna caught in different locations throughout the fish’ natural range. The mean mercury level in the sample was .388 with a std. dev. of 0.12. For the purpose of this lab, assume this is a representative sample.

Note: Since we do not know \(\sigma\) we cannot use the Normal distribution for a CI, we need to use the t-distribution

Question 2

Part A: Describe the parameter of interest in context. State whether we know the value.

# pop. mean mercury level of yellowfin tuna (micro-grams mercury per gram of fish), unknown

Part B: Describe the statistic of interest in context. State whether we know the value.

# sample mean mercury level of yellowfin tuna for 48 fish
# value = .388

Part C: Explain what the conditions are to make a CI using the t-distribution and why they are met in this example.

# We need a random sample and either a Normal population or sample size > 30
# We have a random sample and the sample size n=48 is more than 30.

Part D: How many SE’s do we need to add and subtract to make a 90% CI with our data?

qt(.95, df=47)
## [1] 1.677927

Part E: Make a 90% CI for the parameter. (Show work)

.388 - 1.68 * .12 / sqrt(48); .388 + 1.68 * .12 / sqrt(48)
## [1] 0.3589015
## [1] 0.4170985

Part F: Interpret this confidence interval in context. We should be able to use this interpretation to answer the research question.

# We are 90% confident the pop. mean mercury level of yellowfin tuna is between 0.359 and 0.418

Mercury in Fish Pt. 2

Suppose the 48 fish were actually caught using random samples from 2 different locations: 1) gulf of Mexico and 2) east coast of Japan. We could look at answering a different research question:

Research Question: What is the difference in pop. mean mercury levels for yellowfin tuna in the Gulf of Mexico vs. off the east coast of Japan?

Gulf of Mexico sample

There were 20 fish caught in the gulf of Mexico with a sample mean of 0.413 and std. dev. of 0.15

East Coast of Japan sample

There were 28 fish caught off the east coast of Japan with a sample mean of 0.370 and std. dev. of 0.10

Question 3

Part A: Explain why the conditions to make a 95% CI for the difference in pop. means using a t-distribution are not met for this sample. (regardless, we will continue for practice)

# We need both samples to collected by random sampling (met) and both groups need sample sizes greater than 30, which is not met because the sample sizes are 20 and 28 respectively.

Part B: What is the df for the t-distribution we will use and how many SE’s will we add and subtract for our 95% CI?

df = 19
t = qt(.975, df=19)
t
## [1] 2.093024

Part C: What is the value of the SE?

SE = sqrt(.15^2 / 20 + .1^2 / 28)
SE
## [1] 0.03849861

Part D: Make a 95% CI for the difference in pop. means (keep track of subtraction order)

(.413 - .370) - t*SE
## [1] -0.03757851
(.413 - .370) + t*SE
## [1] 0.1235785

Part E: Interpret the confidence interval in context.

# We are 95% confident that the difference in population mean mercury levels (GoM - Japan) is between 0-.038 and 0.124

Part F: According to the confidence interval, is it plausible there is actually no difference in pop. mean mercury levels?

# Yes. 0 is within the confidence interval, so no difference is plausible according to the 95% CI

Proportions

The dataset below contains the results from a poll based on a random sample with two variables: response, indicating their response to the poll question, and political, reporting their self-reported political ideology.

A number of randomly sampled registered voters from Tampa, FL were asked if they thought workers who have illegally entered the US should be (i) allowed to keep their jobs and apply for US citizenship, (ii) allowed to keep their jobs as temporary guest workers but not allowed to apply for US citizenship, or (iii) lose their jobs and have to leave the country.

## Copy and run this code to create table
immigration <- read.csv("https://collinn.github.io/data/immigrationpoll.csv")

Question 4

We will make a confidence interval to answer the question “What proportion of conservative Tampa voters support workers being ‘allowed to keep their jobs and apply for US citizenship.’

Part A: Describe the parameter of interest, including which symbol we use for it.

# the true population proportion of conservative Tampa, FL voters that support the 'citizenship' option = p

Part B: What is the corresponding value of the statistic and what symbol do we use for it? (making a table may help)

with(immigration, table(response, political)) %>% addmargins(1)
##                        political
## response                conservative liberal moderate
##   Apply for citizenship           57     101      120
##   Guest worker                   121      28      113
##   Leave the country              179      45      126
##   Not sure                        15       1        4
##   Sum                            372     175      363
# p-hat
57 / 372
## [1] 0.1532258

Part C: What is the sample size for the group of conservatives?

372
## [1] 372

Part D: Check the conditions for making a confidence interval.

# Random sample: met
# Success condition: 57 successes (met)
# Failure condition: 315 failures (met)
372 - 57
## [1] 315

Part E: Create a 95% confidence interval for the parameter.

p_hat = (57/352)
n = 372
p_hat - 1.96 * sqrt(p_hat * (1-p_hat) / n)
## [1] 0.1244957
p_hat + 1.96 * sqrt(p_hat * (1-p_hat) / n)
## [1] 0.1993679

Part F: Interpret the confidence interval.

# the true proportion of conservative Tampa, FL voters that support the 'citizenship' option is between 0.12 and 0.20

Question 5

Let’s see if there is a difference between conservatives and liberals in terms of proportions that support workers being ‘allowed to keep their jobs and apply for US citizenship.’

Part A: What is the value of the statistic of interest? (Make ‘Liberal’ the first group)

# proportions, liberal - conservative
(101/175) - (57/372)
## [1] 0.4239171

Part B: Check the conditions to make a confidence interval.

# Random samples (met)
# n1*p1 = 101 > 10 (met)
# n1*(1-p1) = 74 > 10 (met)
# n2*p2 = 52 > 10 (met)
# n2*(1-p2) = 317 > 10 (met)

Part C: Make a 90% confidence interval.

p1 = (101/175)
p2 = (57/372)
n1 = 175
n2 = 372

(p1 - p2) - 1.645 * sqrt(p1*(1-p1)/n1 + p2*(1-p2)/n2)
## [1] 0.3552326
(p1 - p2) + 1.645 * sqrt(p1*(1-p1)/n1 + p2*(1-p2)/n2)
## [1] 0.4926015

Part D: Interpret the confidence interval

# We are 90% confident the difference in proportions of all liberal and all conservative Tampa, FL voters that support the 'citizenship' option is between 0.355 and 0.492.

# equivalently
# We are 90% confident the proportions of all liberal Tampa, FL voters that support the 'citizenship' option is between 0.355 and 0.492 *higher* than the proportion of conservative voters that support this option

Part E: According to the CI, is it plausible there is no difference between the groups?

# No. 0 is nowhere close to the interval.

# According to the CI, a much larger proportion of liberal Tampa voters support the 'citizenship' option compared to conservative voters