Question 1 (Conceptual)

Part A: What is the purpose of a confidence interval?

Part B: What does “95% confidence” actually mean?

Part C: What is the formula for the standard error of the sample mean?

Part D: What is the formula for a 95% confidence interval for a population mean?

Part E: What type of variability are confidence intervals accounting for?

Part F: Do confidence intervals account for biased samples?

Question 2 – Florida Mercury Levels

This is using the info from a study similar to the Florida lakes fish data we saw in the slides.

Data was collected from 107 lakes in Florida. For each lake, the mercury level (ppm) was computed for a large mouth bass. The average mercury level in the sample was 0.554 ppm.

Part A: Describe the parameter in context. What is the corresponding parameter symbol? If we know the value, provide it.

Part B: Describe the statistic in context. State the corresponding statistic symbol. If we know the value, provide it.

Part C: The standard error for this statistic is 0.051. What is the 95% confidence interval? (Show your work)

Part D: Interpret the confidence interval.

Part E: In the U.S., the FDA action level is 1 ppm. Is this safely below the U.S. limit?

Part F: In Canada, the safety limit is 0.5 ppm. Is this clearly above the Canadian limit?

Part G: Suppose we didn’t know the standard error value in part C, but instead we knew the population standard deviation is 0.20. Create a new 95% confidence interval using this info. Does your answer to Part F change?

Part H: What is the practical takeaway from your answer to Part G?

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 3

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

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

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?

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?

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

Part F: Using your answer to C and D, what do you notice about the values we get for the cutoff as the sample size increases? (They should be getting close to a specific value)

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 questions.

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 4

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

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

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

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

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

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

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 5

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)

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?

Part C: What is the value of the SE?

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

Part E: Interpret the confidence interval in context.

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