Two sample t-test quiz

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This accompanies chapter 23 - Comparing two samples from normal distributions

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There are plenty of reasons to choose your partner carefully. In much of the biological world a key reason is “evolutionary fitness” – presumably organisms evolve to choose mates that will help them make more (or healthier) children. This could, for example explain Kermit’s resistance in one of the more complex love stories of our time, as frogs and pigs are unlikely to make healthy children.

To evaluate this this idea @swierk2019, identified a males top choice out of two female wood frogs and then had them mate with the preferred or unpreferred female and counted the number of hatched eggs.

The data can be accessed from the following link: https://raw.githubusercontent.com/ybrandvain/biostat/master/data/Swierk_Langkilde_BEHECO_1.csv

Plot the data: Start with this code. What pattern do you see?

library(dplyr)
library(readr)
library(ggplot2)
frogs <- read_csv("https://raw.githubusercontent.com/ybrandvain/biostat/master/data/Swierk_Langkilde_BEHECO_1.csv")

ggplot(frogs, aes( x = treatment, y = hatched.eggs, ... )) +
geom___() +
stat_summary(fun.data = mean_cl_normal)
*

Code starter

For the next bit's you'll need some code. Feel free to do it all yourself, but if you want a boot, I've got much of it started here

According to the standard interpretation, Cohen's D is

1 point

Small

Medium

Large

Very Large

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From this Cohens D value, we conclude that:

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Reject the null hypothesis

Fail to reject the null hypothesis

The null is false

The null is true

Any actual difference (if it exists) will be subtle (wether significant or not)

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The assumptions of a two-sample t-test are (select all that apply)

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Response variables are normally distributed

Residuals are normally distributed

Variance does not differ between groups

Response variables are independent of predictors

Data within each group is independent

Response variables are collected without bias

Above we saw a difference in variance of 1000 between group (56118 vs 67412), how worried should you be about violating the assumption of homoscedasticity?

1 point

Not very worried

Freaking out

Not enough information

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I wrote this code to generate the qq plots below

From the plot above, I think the data are

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Perfectly normal

Maybe a bit not normal -- I might try a stranformation-- but I'm not freaking out

Freaking out - I need to transform or find another approach

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Say you wanted to transform. Modify this code to makeqq plots for different transformations. Which one best meets normality assumptions

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The original, untransformed data

log10 transformed data: frogs %>% mutate (log10_eggs = log10(hatched.eggs))%>% ggplot(aes(x=...))

exponential transformed data: frogs %>% mutate (exp_eggs = exp(hatched.eggs))%>% ggplot(aes(x=...))

square root transformed data: frogs %>% mutate (sqrt_eggs = sqrt(hatched.eggs))%>% ggplot(aes(x=...))

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Test!

Regardless of your view on the appropriateness of the data for the test. Conduct a classic two-sample t-test on these data using both the lm() and t.test() functions

State the null hypothesis *

State the alternative hypothesis *

Which values are in the 95% confidence interval or the difference (i.e. preferred - preferred) select all that apply

1 point

-92

-42

102

152

202

252

The absolute value of t is about

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0.5

1.0

1.5

2.5

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The p-value is

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2.63e-12

0.304

0.306

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We therefore _____ the null hypothesis

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reject

fail to reject

fail to accept

This means the null hypothesis is

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TRUE

FALSE

?????

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The proportion of variance in eggs hatched explained by treatment is approximately

1 point

Clear selection

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