Null hypothesis: The proportion of people who choose the same hotel again will be the same for beach combers and windsurfers.
To start off, I constructed my data.frame in a very roundabout manner:
>Beachcomber <- c(163,64,227)
>Windsurfer <- c(154,108,262)
>Choose_again <- c("Yes", "No", "Total")
>dat <- data.frame(Choose_again, Beachcomber, Windsurfer)
>dat$Total <- dat$Beachcomber + dat$Windsurfer
>dat <- dat[,-1]
>rownames(dat) <- c("Yes", "no","total")
Beachcomber Windsurfer Total
Yes 163 154 317
no 64 108 172
total 227 262 489
Next, I run the chi-test and save it in an object:
Pearson's Chi-squared test with Yates' continuity correction
data: dat[1:2, 1:2]
X-squared = 8.4903, df = 1, p-value = 0.0035
I was stuck on that for a while, trying to figure out exactly which parts of the data.frame needed to be included/excluded. I did notice that this way of doing it didn’t give me the error that other combinations did though. I assume it’s because the function implemented Yates’ continuity correction, which is automatically implemented for 2×2 tables. The p-value is quite small, so at a threshold of p=.05, the null hypothesis would be rejected and we’d say that it appears that the proportion of people who choose to return to their hotel is different for each group.
gg <- data.frame(x = seq(0,20,.1))
gg$y <- dchisq(gg$x, 1)
geom_ribbon(data=gg[gg$x>qchisq(.05,1,lower.tail=FALSE),], aes(x,ymin=0, ymax=y), fill="red")+
geom_vline(xintercept = res$statistic, color = "blue")+
labs( x = "x", y = "dchisq(gg$x, 1)")+
geom_text(aes(x=8, label="x^2", y=0.25), colour="blue", angle=90)
> high <- c(10,9,8,9,10,8))
> moderate <- c(8,10,6,7,8,8)
> low <- c(4,6,6,4,2,2)
> reaction <- c(10,9,8,9,10,8,8,10,6,7,8,8,4,6,6,4,2,2)
> stresslvls <- c(rep("high",6), rep("moderate",6), rep("low",6))
> dat <- data.frame(reaction,stresslvls)
> analysis <- lm(reaction ~ as.factor(stresslvls), data = dat)
Analysis of Variance Table
Df Sum Sq Mean Sq F value Pr(>F)
as.factor(stresslvls) 2 82.111 41.056 21.358 4.082e-05 ***
Residuals 15 28.833 1.922
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The ANOVA test essentially breaks a sample down into smaller groups and compares the variance between the means of each of those groups to determine how much variance can be attributed to chance.
Here, we have analyzed a grouping of the reaction times of participants that were subjected to various levels of stress. Above, we can see the return of the ANOVA test. We can see that the between-group mean sum of squares is 41.056. This shows that the difference between each of the individual sample means and the mean of all the samples. The F value is the ratio between the mean squares of the mean squares. The larger the F statistic is, the more you can rule out the difference between the means being due to chance. The fact that the P value is so small allows us to reject the null hypothesis and claim that the level of stress the participants are subjected does have an effect on their reaction times.
1a. H0: A cookie will break at more than 70 pounds of force or greater; u >= 70
Ha: A cookie will break at less than 70 pounds of force on average; u < 70
1c. p = 0.0359 There’s a 3.59% chance of finding a sample mean of 69.1 if u >=70.
1d. p = 0.0002 since p < 0.5, there is enough evidence to dismiss H0
1e. p = 0.0227 since p < 0.5, there is enough evidence to dismiss H0
2a. cor(mod8$Cost.per.serving, mod8$Fibber.per.serving)
Q1. I created a function to do the calculations to find mu for the Confidence Interval:
ci <- function(xbar, con, sig, n)
con1 = con + ((1-con)/2)
mu = xbar + (qnorm(con1) * (sig / sqrt(n)))
sn = xbar – (qnorm(con1) * (sig / sqrt(n)))
paste(“The confidence interval is between “, sn, ” and “, +mu, ” and the estimated mean is “, xbar, sep =””)
The confidence intervals came in as follows:
> ci(115, .9, 15, 100)
 “The confidence interval is between 112.532719559573 and 117.467280440427 and the estimated mean is 115”
> ci(115, .95, 15, 100)
 “The confidence interval is between 112.06005402319 and 117.93994597681 and the estimated mean is 115”
> ci(115, .98, 15, 100)
 “The confidence interval is between 111.510478188939 and 118.489521811061 and the estimated mean is 115”
 “The confidence interval is between 83.0400360154599 and 86.9599639845401 and the estimated mean is 85”
((1.96 * 15) / 5)^2
I think the minimum sample size should be 35 if I’m thinking of the zscore tails properly.
edit: After reviewing the last question, 35 is right, but how i got to it was wrong
The biggest thing to note about this is the difference between the 95% confidence interval and the 97.5th percentile ranking. Each tail of the 95% confidence interval contains 2.5% of the area under the curve. But we’re only looking for a single tailed value, so we must use qnorm(.975).
I feel like I was perhaps a little bit confused about what questions 1 and 2 were asking for, or what the answer should look like.
Import vector of values of ice cream purchase numbers
>a <- c(8, 14, 16, 10, 11)
Generate random sample of 2 values and save to a vector
> b <- sample(a,2)
 10 11
Calculate mean and standard deviation of the sample.
Create data.frame out of mean and stdev for the sample and population.
smp <- c(10.5,0.7071068)
> pop <- c(11.8,3.193744)
> c <- data.frame < (smp, pop)
- I think that the sample proportion will have the approximately the same distribution since nq = 5.
- I think 100 is the smallest value of n for which p is approximately normal because anything smaller than n = 100 will make np < 5. The high value of p is very limiting.
Initially, I was just going to calculate the correlation coefficient manually in R without using the functions, but then I decided that if I was going to do it manually then I may as well do it by hand. I ended up making a mistake with spacing by not thinking about how many digits would be made out of squaring an already triple digit number.
From there, inputting the x and y values into R and executing the cor.tests() were very simple and straightforward. I feel like I’ve gotten to the point with R where I don’t feel like I’m completely lost all the time.
For this assignment, I decided to to make everything as one script and use the Source with Echo command to run it all at once. The only issue I had is that mode() does not yield a mode. It returns what storage mode the object uses, so I had to look up how to make a function to calculate the mode.
From there, I was able to call mat and matrange to pull up the tables of all the values.
The process of setting up git with RStudio was not as straight-forward as I had expected. I had not realized that git was different from the Github desktop app. Trying to tie the Github desktop app into R Studio caused R Studio to keep opening up new instances of the Github desktop app while it tried to establish a connection to the website. After realizing that there was, in fact, a difference between git and the desktop app, I refocused my efforts on git. After redoing all the configuration, I tried to push my package to github only for it to fail again. Somewhere along the line of my configurations, the actual repository itself didn’t get created.
For the package itself, I tried to make a simple series of functions to do some goat census work for a game I play. It didn’t go well at all. I’m currently taking java, so I’m having some challenges switching back and forth between syntaxes, causing some turbulence with creating some simple loops. I spent an embarrassing amount of time trying to figure out why my loop wasn’t working (I do know that loops aren’t common in R). It turns out that ++i doesn’t work. In the midst of me trying to work that out, I was trying to have my functions stand alone in their own files to be called from the main file, and that wasn’t working at all either. I tried various source declarations and, in the end, wasn’t able to get it to work. All of this troubleshooting left me with a sad little shriveled raisin of a package.