The 95% confidence interval is (0.1739, 0.9118). Therefore, I am 95% confident that the average height of treated radishes is greater than the average height of untreated radishes by between 0.1739 and 0.9118 cm.
My confidence interval fails to support this claim because the bounds of my interval both lie below this claim, therefore 1 cm is not in my confidence interval, and cannot not be supported by this evidence.
The largest difference in growth I could expect to see with 95% confidence, based on my interval would be 0.9118. This would be the largest difference between how much taller the average of the treated plants are compared to the controlled.
The 99% confidence interval is (9.3745, 12.4827). So, I am 99% confident that the difference between chirps with music and chirps without music is between 9.3745 and 12.4827.
The test statistic for a two-sided hypothesis is 19.4834649, with corresponding p-value 1.975478710^{-17}.
The test statistic for the alternative \(H_A: \mu_1 > \mu_2\) is 19.4834649, with corresponding p-value 9.877393410^{-18}.
The test statistic for the null \(H_O: \mu_1 ≤ \mu_2\) is 19.4834649, with corresponding p-value 1.
The p-value 1.975478710^{-17} is less than our α 0.01, so I can reject the null, and therefore proving that with 1% significance, the average amount of chirps of the birds with music is higher than the average without chirps.
I think it makes sense in most cases to compare the confidence interval and the hypothesis test. Even though they give us pretty similar results, the confidence interval just gives you a difference, but the hypothesis test tells you whether it is significant. If we had a larger variance of numbers, the differences may not precisely compute to being significant. This case it may not have been necessary, but it does help in coming to a conclusion.
The 95% confidence interval is (-18.6759, -1.233). Therefore, I am 95% confident that the average score of untreated students is less than the average score of treated students by between -18.6759 and -1.233.
The test statistic for a two-sided hypothesis is -2.3108892, with corresponding p-value 0.0263824.
The test statistic for the null \(H_O: \mu_1 ≥ \mu_2\) is -2.3108892, with corresponding p-value 0.9868088.
The test statistic for the alternative \(H_A: \mu_1 < \mu_2\) is -2.3108892, with corresponding p-value 0.0131912.
The p-value 0.0263824 is less than our α 0.05, so I can reject the null, and therefore proving that with 5% significance, the average score of the treated students is higher than the average without treatment.
Yes, my conclusion would have changed if the α=0.01 because the p-value was about 0.026. Therefore the p-value would have been greater than the α, and I would have then failed to reject the null. I would then come to the opposite conclusion than when α=0.05.
Radish=read.csv("~/Desktop/STA 032 - R HW 1_files/Radish.csv")
the.stuff = t.test(Height ~ Treatment, data = Radish, conf.level = 0.95)
the.CI = round(the.stuff$conf.int,digits = 4)
Chirp=read.csv("~/Desktop/STA 032 - R HW 1_files/Chirp.csv")
the.stuffc = t.test(Chirp$With, Chirp$Without, paired = TRUE, data = Chirp, conf.level = 0.99)
the.CIc = round(the.stuffc$conf.int,digits = 4)
two.side = t.test(Chirp$With, Chirp$Without, paired = TRUE, conf.level = 0.99,alternative = "two.sided")
greater = t.test(Chirp$With, Chirp$Without, paired = TRUE, conf.level = 0.99,alternative = "greater")
less = t.test(Chirp$With, Chirp$Without, paired = TRUE, conf.level = 0.99,alternative = "less")
DRP=read.csv("~/Desktop/STA 032 - R HW 1_files/DRP.csv")
the.stuffd = t.test(Response ~ Treatment, data = DRP, conf.level = 0.95)
the.CId = round(the.stuffd$conf.int,digits = 4)
two.sideb = t.test(Response ~ Treatment, data = DRP, conf.level = 0.95,alternative = "two.sided")
greaterb = t.test(Response ~ Treatment, data = DRP, conf.level = 0.95,alternative = "greater")
lessb = t.test(Response ~ Treatment, data = DRP, conf.level = 0.95,alternative = "less")