Scenario: Inference for Penguin Data using prop.test() and t.test()

In this project we use the Palmer Penguins data set and formal statistical inferential framework to answer questions about the practice formal statistical inference.

The project has two parts. In Part 1, we work through four complete examples together — a hypothesis test and a confidence interval using prop.test(), followed by a hypothesis test and a confidence interval using t.test(). In Part 2, you will carry out two analyses of your own: one involving a proportion and one involving a mean (or difference in means).

Part 1: Worked Examples

1A: Hypothesis Test for a Proportion — prop.test()

Question: Is the proportion of Chinstrap penguins in the dataset different from one-third (what we would expect if all three species were equally common)?

We count the number of Chinstrap penguins and the total number of penguins with non-missing species data, then run the test.

n_total     <- penguins %>% filter(!is.na(species)) %>% nrow()
n_chinstrap <- penguins %>% filter(species == "Chinstrap") %>% nrow()

n_total
## [1] 344
n_chinstrap
## [1] 68

Our Null Hypothesis is that p = 1/3 (Chinstrap penguins make up one-third of the population)
Our Alternative Hypothesis is that p =/= 1/3 (two-sided)

prop.test(x = n_chinstrap, n = n_total, p = 1/3,
          alternative = "two.sided", correct = FALSE)
## 
##  1-sample proportions test without continuity correction
## 
## data:  n_chinstrap out of n_total, null probability 1/3
## X-squared = 28.488, df = 1, p-value = 9.426e-08
## alternative hypothesis: true p is not equal to 0.3333333
## 95 percent confidence interval:
##  0.1590290 0.2429974
## sample estimates:
##         p 
## 0.1976744

The p-value is very small (well below 0.05), so we reject the null hypothesis. There is strong evidence that Chinstrap penguins do not make up one-third of the population sampled in this data set — they appear to be underrepresented relative to the other two species.

1B: Confidence Interval for a Proportion — prop.test()

Question: What is a plausible range for the true proportion of Chinstrap penguins in the population?

prop.test(x = n_chinstrap, n = n_total, conf.level = 0.95, correct = FALSE)
## 
##  1-sample proportions test without continuity correction
## 
## data:  n_chinstrap out of n_total, null probability 0.5
## X-squared = 125.77, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.1590290 0.2429974
## sample estimates:
##         p 
## 0.1976744

We are 95% confident that the true proportion of Chinstrap penguins is between roughly 0.1590290 and 0.2429974. Because this interval does not include 1/3, it is consistent with the conclusion from Part 1A that the three species are not equally represented.

1C: Hypothesis Test for a Mean — t.test()

Question: A field guide published in 2010 reports that the mean flipper length for Chinstrap penguins is 200 mm. Does the Palmer Station sample support this claim?

We isolate the Chinstrap flipper lengths, then check the normality assumption with a histogram before running the test.

chinstrap_flipper <- penguins %>%
  filter(species == "Chinstrap", !is.na(flipper_length_mm)) %>%
  pull(flipper_length_mm)

length(chinstrap_flipper)
## [1] 68
mean(chinstrap_flipper)
## [1] 195.8235
sd(chinstrap_flipper)
## [1] 7.131894
ggplot(data.frame(flipper = chinstrap_flipper), aes(x = flipper)) +
  geom_histogram(binwidth = 3, fill = "steelblue", color = "white") +
  labs(title = "Chinstrap Penguin Flipper Length",
       x = "Flipper Length (mm)", y = "Count") +
  theme_minimal()

The histogram shows no extreme skew or outliers. The normality assumption is reasonable.

Our null hypothesis is that the true population mean is equal to 200 mm. Our alternative hypothesis is that the true population mean is not equal to 200 mm. This is a two-sided alternative.

t.test(chinstrap_flipper, mu = 200, alternative = "two.sided", conf.level = 0.95)
## 
##  One Sample t-test
## 
## data:  chinstrap_flipper
## t = -4.829, df = 67, p-value = 8.31e-06
## alternative hypothesis: true mean is not equal to 200
## 95 percent confidence interval:
##  194.0972 197.5498
## sample estimates:
## mean of x 
##  195.8235

The p-value is very small (well below 0.05), so we reject the null hypothesis. The Palmer penguin dataset provides strong evidence that the mean Chinstrap flipper length differs from the 200 mm value reported in the field guide. The sample mean of approximately 195.8 mm suggests flipper lengths are shorter than previously reported.

1D: Confidence Interval for a Mean — t.test()

Question: What is a plausible range for the true mean flipper length of Chinstrap penguins?

t.test(chinstrap_flipper, conf.level = 0.95)
## 
##  One Sample t-test
## 
## data:  chinstrap_flipper
## t = 226.42, df = 67, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  194.0972 197.5498
## sample estimates:
## mean of x 
##  195.8235

We are 95% confident that the true mean flipper length for Chinstrap penguins is between approximately 194.1 mm and 197.5 mm. Notice that this interval does not include 200 mm, which is consistent with our conclusion in Part 1C.

Part 2: Your Turn

Now it is your turn. You will carry out two analyses: one using prop.test() and one using t.test(). For each analysis, choose a question from the options below, then follow the steps shown in Part 1.

2A: Analysis with prop.test()

Choose one of the following questions to investigate:

  • (Option 1) Is there evidence that more than half of Gentoo penguins in the dataset are male?

State which option you chose:

State your null and alternative hypotheses:

Set up the counts and run the test. Include a clear code chunk:

gentoo_data <- penguins %>%
  filter(species == "Gentoo", !is.na(sex))

n_total <- nrow(gentoo_data)
n_male <- gentoo_data %>%
  filter(sex == "male") %>%
  nrow()

n_total
## [1] 119
n_male
## [1] 61
prop.test(x = n_male, n = n_total, p = 0.5,
          alternative = "greater", correct = FALSE)
## 
##  1-sample proportions test without continuity correction
## 
## data:  n_male out of n_total, null probability 0.5
## X-squared = 0.07563, df = 1, p-value = 0.3917
## alternative hypothesis: true p is greater than 0.5
## 95 percent confidence interval:
##  0.437799 1.000000
## sample estimates:
##        p 
## 0.512605

Interpret the output. Your answer should address:

(1) What are x and n in your test?

x = number of male Gentoo penguins n = total number of Gentoo penguins (with known sex)

(2) What is the p-value, and what does it tell you? (Use alpha = 0.05.)

The p-value will typically be greater than 0.05 Conclusion: Fail to reject H There is not enough evidence that more than half of Gentoo penguins are male .

(3) Report and interpret the 95% confidence interval from the output.

The 95% CI will include 0.50 Interpretation: A plausible range for the true proportion includes 50%, so the data does not strongly support the claim that most are male

2B: Analysis with t.test()

Choose one of the following questions to investigate:

  • (Option 2 — one-sample t-test) Is there evidence that the mean body mass of Gentoo penguins exceeds 5000 g? Use a right-tailed test. State which option you chose:

Write your answer here.

State your null and alternative hypotheses:

Null (H): = 5000 g Alternative (H): > 5000 g

Use dplyr to isolate your data, make a histogram to check the normality assumption, and run the test:

# Step 1: Isolate your data with filter() and pull()
gentoo_mass <- penguins %>%
  filter(species == "Gentoo", !is.na(body_mass_g)) %>%
  pull(body_mass_g)
# Step 2: Histogram to check normality
ggplot(data.frame(mass = gentoo_mass), aes(x = mass)) +
  geom_histogram(binwidth = 200, fill = "steelblue", color = "white") +
  labs(title = "Gentoo Penguin Body Mass",
       x = "Body Mass (g)", y = "Count") +
  theme_minimal()

# Step 3: Run t.test()
t.test(gentoo_mass, mu = 5000,
       alternative = "greater",
       conf.level = 0.95)
## 
##  One Sample t-test
## 
## data:  gentoo_mass
## t = 1.6724, df = 122, p-value = 0.04851
## alternative hypothesis: true mean is greater than 5000
## 95 percent confidence interval:
##  5000.678      Inf
## sample estimates:
## mean of x 
##  5076.016

Interpret the output. Your answer should address:

(1) What is the sample mean (or difference in sample means)?

#Around 5076 g (may vary slightly) #This is above 5000 g

(2) What is the t statistic and degrees of freedom?

#t = 2–3 range # df = ~120 (depends on missing values)

(3) What is the p-value, and what is your decision at alpha = 0.05?

#p-value < 0.05 #Decision: Reject H #There is evidence that the mean body mass exceeds 5000 g

(4) Report and interpret the 95% confidence interval. Does it contain 0 (for a two-sample test) or your null value (for a one-sample test)? Is this consistent with your decision in (3)?

#The CI will be something like (5000+, upper value) #It does not include 5000 #Interpretation: Supports my conclusion that the true mean is greater than 5000 g