Problem 1:

Swearing, the use of offensive or obscene language occurs, in most human cultures. People swear to let off steam, to shock or insult, or out of habit. Cathartic swearing may occur in painful situations, for example, giving birth or hitting one’s thumb with a hammer. To test whether swearing increases one’s tolerance for pain the following experiment was conducted. Twenty college-aged men were recruited for the study. At the outset they were informed that the study was concerned with quantifying the degree of stress that various forms of language elicit during tense situations. Participants submerged their non-dominant hand in room temperature water for three minutes and then immersed the same hand in 5◦C water with the instruction that they should submerge their unclenched hand for as long as possible while repeating a common word, for example, “chair” or a swear word of their choice. The order of the two trials was determined by a coin flip. The data for these 20 participants with durations in 5◦C water recorded in seconds is contained in the csv file (HW3).

a) Conduct a paired t-test on the data. Report the P value and the confidence interval for the mean of the difference column. Explain results.

test <- t.test(swears$Swearing,
               swears$`Not Swearing`,
               paired = TRUE)
test
## 
##  Paired t-test
## 
## data:  swears$Swearing and swears$`Not Swearing`
## t = 4.0503, df = 19, p-value = 0.000683
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  15.24641 47.85359
## sample estimates:
## mean of the differences 
##                   31.55

p-value = 0.000683 < 0.05

Confidence Interval= (15.24641, 47.85359)

Null Hypothesis: the mean of the hypothetical difference column = 0 (we didn’t actually make a difference column). In other words, there is no difference between the two treatments (swearing vs not swearing)

Because the p-value < 0.05, we reject the null hypothesis that there is no mean difference between the treatments. The confidence interval further supports that there is a significant difference between treatment means because it does not contain 0; we are 95% confident that swearing, on average, makes a difference in pain tolerance.

b) Do a correlation analysis to assess how effective is the pairing (see page 311 in the book). Explain the results.

cor <-cor.test(swears$Swearing,
               swears$`Not Swearing`,
               method=c("pearson"))
cor
## 
##  Pearson's product-moment correlation
## 
## data:  swears$Swearing and swears$`Not Swearing`
## t = 7.1716, df = 18, p-value = 1.121e-06
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.6753765 0.9437502
## sample estimates:
##       cor 
## 0.8606703

Confidence Interval = (0.6753765 0.9437502)

correlation coefficient(r) = 0.8606703

coefficient of determination (r^2) = 0.740753365

p-value = 1.121e-06

The correlation coefficient (r) is positive but <1, so the two values are positively correlated but scattered. The coefficient of determination, r^2 = 0.74, meaning that 74% of the variability can be explained by swearing or not swearing during the ice water test. The remaining 26% would be explained by unrelated outside factors or errors. This correlation is fairly strong and shows us that the pairing is effective, especially since the paired p-value (part a) is much smaller than the unpaired p-value (part c). Additionally, the p-value for r is < 0.05 to further support this.

c) Conduct an unpaired t-test (yes! again) with the same dataset assuming two independent samples with equal variance. Report the P value and the confidence interval for the difference between means. Compare the results with part “a.”

sweartest <- t.test(swears$Swearing,
                    swears$`Not Swearing`,
                    paired = FALSE, var.equal = TRUE)
sweartest
## 
##  Two Sample t-test
## 
## data:  swears$Swearing and swears$`Not Swearing`
## t = 1.7034, df = 38, p-value = 0.09667
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -5.946296 69.046296
## sample estimates:
## mean of x mean of y 
##    195.95    164.40

p value = 0.09667 > 0.05

confidence interval = ( -5.946296 69.046296)

Null Hypothesis: There is no significant difference between the means two treatment groups.

Given a p-value >0.05 and confidence intervals that include 0, we cannot reject the null hypothesis– there is no significant difference between the two treatment means. In this case, it was worth it to do a paired t-test (part a) in order to preserve statistical power because the correlation analysis showed a strong positive relationship between swearing and not swearing treatments. This makes sense, because the experimental design was established as paired–each individual participated in both the swearing and not swearing treatment.

d) Conduct a linear regression analysis equivalent to the unpaired t-test (see page 361 in the book). Report the slope with its confidence interval, and the P value based on the F ratio test. Explain the results. NOTE: the p-value should be identical to the one obtained in part “c.”

binaryswears <- read_excel("~/Documents/Biometry/binaryswearsHW3.xlsx")
swear_regression <- lm (Time ~ Swearornot, data=binaryswears)
summary(swear_regression)
## 
## Call:
## lm(formula = Time ~ Swearornot, data = binaryswears)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -118.95  -45.20    9.55   43.55   95.05 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   164.40      13.10  12.552 4.32e-15 ***
## Swearornot     31.55      18.52   1.703   0.0967 .  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 58.57 on 38 degrees of freedom
## Multiple R-squared:  0.07094,    Adjusted R-squared:  0.04649 
## F-statistic: 2.901 on 1 and 38 DF,  p-value: 0.09667
confint(swear_regression, 'Swearornot', level = 0.95)
##                2.5 %  97.5 %
## Swearornot -5.946296 69.0463

Slope = 31.55, CI of slope = (-5.946296 69.0463) Intercept = 164.4

p-value = 0.09667

Null hypothesis: There is no linear relationship between swearing and not swearing treatments.

The slope represents the difference between the means still, because we are working with binary data (swearing = 1, no swearing = 0).The p-value for the slope > 0.05 and confidence intervals do contain 0, so we cannot reject the null hypothesis. This does necessarily mean that there is NO relationship between swearing and not swearing, simply that the relationship, if it exists, is not linear. Because the p-value (=0.09667) for the F-ratio > 0.05, we cannot reject the null that the relationshi between the two treatments are a straight line (no difference between treatment means)