Problem I

(a)

The CI is (0.1739 , 0.9118)

This means, in plain English, that we are 95% confident that the true average growth of treated radishes is between 0.1739 and 0.9118 cms greater than that of the control group.

(b)

My confidence interval does NOT support the claim, as the confidence interval does not contain ‘1 cm’.

(c)

The largest difference in growth I can expect with 95% confidence is that the treated plants will be .9118 cms taller, as they were group one in r and the confidence interval values were both positive, meaning that their values are greater than those of group two (the control group).

Problem II

(a)

## 
##  Paired t-test
## 
## data:  chirp$With and chirp$Without
## t = 19.483, df = 27, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 99 percent confidence interval:
##   9.374453 12.482690
## sample estimates:
## mean of the differences 
##                10.92857

This means, in plain English, that we are 99% confident that the true average difference in number of chirps while the birds are listening to music is between 9.37 and 12.48 greater than when the birds are not listening to music.

(b)

H_0: mu_D equals 0. H_A: _D does not equal 0.

So that the 99% CI is: (9.0315,12.8257), with corresponding test-statistic 15.6351, and p-value 0.

(c)

The p-value means that, with 99% confidence, we can say that there is a 0% chance of observing our test statistic or a more extreme one if H_0 is true. At ?? = 0.01, we reject the null.

(d)

Yes, because while the confidence interval would lead us to believe the data is correct, the hypothesis says otherwise. And comparing them leads me to believe that there is an error.

Problem III

(a)

## 
##  Welch Two Sample t-test
## 
## data:  Response by Treatment
## t = -2.3109, df = 37.855, p-value = 0.02638
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -18.67588  -1.23302
## sample estimates:
## mean in group Control mean in group Treated 
##              41.52174              51.47619

This means that we are 95% confident that the true average difference in the students’ DRP scores is between 1.23302 and 18.67588 less in the control group than in the treated group.

(b)

H_0: mu_D is less than or equal to. H_A: _D is greater than 0.

So that the 95% CI is: (-17.2176,), with corresponding test-statistic -2.3109, and p-value 0.9868.

(c)

The p-value means that, with 95% confidence, we can say that there is a 98.68% chance of observing our test statistic or a more extreme one if H_0 is true. At ?? = 0.05, we fail to reject the null.

(d)

Yes my conclusion would have changed if I had been using ?? = 0.01, because my p-value would have had to exceed or match 99% in order to fail to reject the null.

Code

Problem I

radish <- read.csv("C:/Users/Lisa/Downloads/Radish.csv")

(a)

the.stuff = t.test(Height ~ Treatment, data = radish, conf.level = 0.95)
the.CI = round(the.stuff$conf.int,digits = 4)

The CI is (0.1739 , 0.9118)

This means, in plain English, that we are 95% confident that the true average growth of treated radishes is between 0.1739 and 0.9118 cms greater than that of the control group.

(b)

My confidence interval does NOT support the claim, as the confidence interval does not contain ‘1 cm’.

(c)

The largest difference in growth I can expect with 95% confidence is that the treated plants will be .9118 cms taller, as they were group one in r and the confidence interval values were both positive, meaning that their values are greater than those of group two (the control group).

Problem II

chirp <- read.csv("C:/Users/Lisa/Downloads/Chirp.csv")

(a)

t.test(chirp$With, chirp$Without, paired = TRUE, conf.level = 0.99, alternative = "two.sided")
## 
##  Paired t-test
## 
## data:  chirp$With and chirp$Without
## t = 19.483, df = 27, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 99 percent confidence interval:
##   9.374453 12.482690
## sample estimates:
## mean of the differences 
##                10.92857

This means, in plain English, that we are 99% confident that the true average difference in number of chirps while the birds are listening to music is between 9.37 and 12.48 greater than when the birds are not listening to music.

(b)

H_0: mu_D equals 0. H_A: _D does not equal 0.

paired.two.side = t.test(chirp$With, chirp$Without, data = chirp, conf.level = 0.99, alternative = "two.sided")
the.CI = round(paired.two.side$conf.int,digits=4)
the.statistic = round(paired.two.side$statistic, digits = 4)
the.pval = round(paired.two.side$p.value,digits = 4)

So that the 99% CI is: (9.0315,12.8257), with corresponding test-statistic 15.6351, and p-value 0.

(c)

The p-value means that, with 99% confidence, we can say that there is a 0% chance of observing our test statistic or a more extreme one if H_0 is true. At ?? = 0.01, we reject the null.

(d)

Yes, because while the confidence interval would lead us to believe the data is correct, the hypothesis says otherwise. And comparing them leads me to believe that there is an error.

Problem III

DRP <- read.csv("C:/Users/Lisa/Downloads/DRP.csv")

(a)

t.test(Response ~ Treatment, data = DRP, conf.int = 0.95, alternative = "two.sided")
## 
##  Welch Two Sample t-test
## 
## data:  Response by Treatment
## t = -2.3109, df = 37.855, p-value = 0.02638
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -18.67588  -1.23302
## sample estimates:
## mean in group Control mean in group Treated 
##              41.52174              51.47619

This means that we are 95% confident that the true average difference in the students’ DRP scores is between 1.23302 and 18.67588 less in the control group than in the treated group.

(b)

H_0: mu_D is less than or equal to. H_A: _D is greater than 0.

paired.greater = t.test(Response ~ Treatment, data = DRP, conf.level = 0.95,alternative = "greater")
the.CI = round(paired.greater$conf.int,digits=4)
the.statistic = round(paired.greater$statistic, digits = 4)
the.pval = round(paired.greater$p.value,digits = 4)

So that the 95% CI is: (-17.2176,), with corresponding test-statistic -2.3109, and p-value 0.9868.

(c)

The p-value means that, with 95% confidence, we can say that there is a 98.68% chance of observing our test statistic or a more extreme one if H_0 is true. At ?? = 0.05, we fail to reject the null.

(d)

Yes my conclusion would have changed if I had been using ?? = 0.01, because my p-value would have had to exceed or match 99% in order to fail to reject the null.