Describe the Central Limit Theorem and also include why it’s appropriate to use a symmetric confidence interval.
Consider the variable LS (litter size) from the carnivora dataset. The output of a t-test is below. Answer the following question based on this output.
carnivora <- read.csv("https://raw.githubusercontent.com/roualdes/data/refs/heads/master/carnivora.csv")
t.test(carnivora$LS, mu = 2, conf.level = 0.9)
One Sample t-test
data: carnivora$LS
t = 9.87, df = 109, p-value < 2.2e-16
alternative hypothesis: true mean is not equal to 2
90 percent confidence interval:
3.024773 3.438863
sample estimates:
mean of x
3.231818 How many observations were measured / what is the sample size?
Interpret, in context of these data, the confidence interval.
Did this confidence interval capture the true population parameter it is targeting?
If we had instead more observations in the sample, would the confidence interval be narrower or wider? Explain why, using concepts and keywords from this class.
If we instead made a 98% confidence interval, would the confidence interval be narrower or wider? Explain why, using concepts and keywords from this class.
Consider again the carnivora dataset, where GL represents gestation length in days and WA represents weaning age in days. The output of a t-test is below. Answer the following questions based on this output.
carnivora$diff <- carnivora$GL - carnivora$WA
t.test(carnivora$diff, conf.level = 0.99)
One Sample t-test
data: carnivora$diff
t = -2.5586, df = 57, p-value = 0.01319
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
-69.848307 1.420721
sample estimates:
mean of x
-34.21379 What type of hypothesis test is this, what is the proper name?
Write the corresponding null and alternative hypotheses using proper statistical symbols.
What is the value for the level of significance?
Write a statistical conclusion using the p-value.
Write an interpretation of the conclusion from d. with minimal statistical jargon.
Explain why the confidence interval supports the same conclusion as the p-value.
Consider again the carnivora dataset. The output of a t-test is below. Answer the following questions based on this output.
t.test(LS ~ SuperFamily, data = carnivora, conf.level = 0.9)
Welch Two Sample t-test
data: LS by SuperFamily
t = 4.4126, df = 76.408, p-value = 3.306e-05
alternative hypothesis: true difference in means between group Caniformia and group Feliformia is not equal to 0
90 percent confidence interval:
0.620908 1.373465
sample estimates:
mean in group Caniformia mean in group Feliformia
3.712281 2.715094 What type of hypothesis test is this, what is the proper name?
Write the corresponding null and alternative hypotheses using proper statistical symbols.
What does this output tell you about these data? Write a conclusion with little to no statistical jargon, citing the hypothesis test or the confidence interval to support your claim.
Consider again the carnivora dataset, where BW represents birth weight in grams. Based on the following code, answer the questions below.
ggplot(carnivora, aes(Family, BW)) + geom_boxplot()Warning: Removed 50 rows containing non-finite outside the scale range
(`stat_boxplot()`).
fit <- lm(BW ~ Family, data = carnivora)
anova(fit)Analysis of Variance Table
Response: BW
Df Sum Sq Mean Sq F value Pr(>F)
Family 7 3256952 465279 6.142 2.648e-05 ***
Residuals 54 4090670 75753
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1TukeyHSD(aov(fit)) Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = fit)
$Family
diff lwr upr p adj
Canidae-Ailuridae 101.68182 -804.63402 1007.99765 0.9999624
Felidae-Ailuridae 313.26250 -581.17450 1207.69950 0.9531564
Hyaenidae-Ailuridae 991.70000 -71.04952 2054.44952 0.0839689
Mustelidae-Ailuridae -47.33933 -943.52842 848.84975 0.9999998
Procyonidae-Ailuridae 6.30000 -963.85314 976.45314 1.0000000
Ursidae-Ailuridae 537.40000 -464.56985 1539.36985 0.6928677
Viverridae-Ailuridae -8.20000 -918.28431 901.88431 1.0000000
Felidae-Canidae 211.58068 -128.28776 551.44912 0.5157207
Hyaenidae-Canidae 890.01818 222.98779 1557.04857 0.0023455
Mustelidae-Canidae -149.02115 -493.47417 195.43186 0.8692106
Procyonidae-Canidae -95.38182 -602.02777 411.26413 0.9988257
Ursidae-Canidae 435.71818 -129.46904 1000.90540 0.2480542
Viverridae-Canidae -109.88182 -489.02093 269.25730 0.9834466
Hyaenidae-Felidae 678.43750 27.63899 1329.23601 0.0352976
Mustelidae-Felidae -360.60183 -672.46244 -48.74123 0.0130218
Procyonidae-Felidae -306.96250 -792.03907 178.11407 0.4946297
Ursidae-Felidae 224.13750 -321.79817 770.07317 0.8970179
Viverridae-Felidae -321.46250 -671.25619 28.33119 0.0932343
Mustelidae-Hyaenidae -1039.03933 -1692.24376 -385.83491 0.0001564
Procyonidae-Hyaenidae -985.40000 -1736.87739 -233.92261 0.0029540
Ursidae-Hyaenidae -454.30000 -1246.42672 337.82672 0.6174485
Viverridae-Hyaenidae -999.90000 -1672.04181 -327.75819 0.0004772
Procyonidae-Mustelidae 53.63933 -434.66037 541.93904 0.9999674
Ursidae-Mustelidae 584.73933 35.93784 1133.54082 0.0290616
Viverridae-Mustelidae 39.13933 -315.11051 393.38917 0.9999661
Ursidae-Procyonidae 531.10000 -131.64076 1193.84076 0.2066827
Viverridae-Procyonidae -14.50000 -527.85679 498.85679 1.0000000
Viverridae-Ursidae -545.60000 -1116.81070 25.61070 0.0710874Is the group degrees of freedom correct? Explain.
What is the appropriate statistical conclusion from this test?
Write an interpretation of the conclusion with minimal statistical jargon.
Explain why it is appropriate to use Tukey’s HSD here.
What does Tukey’s HSD tell you about these possums?