Math411 Key_Lab 2

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1/23/24, 9:32 PM Math411 Key_Lab 2 file:///Users/zacharykey/Math411-Key_Lab-1.html 1/10 Math411 Key_Lab 2 2024-01-17 T-test for Two Independent Random Samples Exercise 1 #A Null Hypothesis: The population length mean in small and large bass are equal. Alternative Hypothesis: The population length mean in small and large bass are not equal. Assumptions: 1. Samples are collected randomly from each of the two populations of interest. 2.The variable is measured on a continuous scale 3.The variable is approximately normally distributed #B #two sample t-test mean1=272.8 mean2=164.8 sd1=96.4 sd2=40 n1=125 n2=97 ttest=(mean1-mean2)/sqrt(((sd1*sd1)/n1)+(sd2*sd2)/n2) ttest ## [1] 11.33153 tcrit=qt(.05/2,sd2-1,lower.tail = F) tcrit ## [1] 2.022691 The t-test gave a value of 11.33 while the t-critical value come out to be 2.023. Since the t value is greater than the t-critical value we would reject the null. This means that the population lengths of the large and small bass are probability not equal.

1/23/24, 9:32 PM Math411 Key_Lab 2 file:///Users/zacharykey/Math411-Key_Lab-1.html 2/10 #C lb=(mean1-mean2)-tcrit*sqrt((sd1^2/n1)+(sd2^2/n2)) ub=(mean1-mean2)+tcrit*sqrt((sd1^2/n1)+(sd2^2/n2)) bin=c(lb,ub) bin ## [1] 88.72189 127.27811 #D In part B the comparison between the t-value and t-critical value supported that the population mean lengths were not equal to each other. In part C we were to find the 95% confidence interval in the difference of the means. If they were equal the values should have been zero but in this case the difference ranged from 89.23 to 126.76. #E largebass=rnorm(n=n1,mean=mean1,sd=sd1) smallbass=rnorm(n=n2,mean=mean2,sd=sd2) t.test(largebass,smallbass,varequal=T) ## ## Welch Two Sample t-test ## ## data: largebass and smallbass ## t = 11.552, df = 181.32, p-value < 2.2e-16 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## 93.39527 131.86998 ## sample estimates: ## mean of x mean of y ## 274.2540 161.6213 The output of the random sample of each population when run through a t-test gave us a t-value of 11.728 and a p-value less than 2.2e-16. The p-value being less than our critical value 0.05 supports the rejection of the null.This shows that even in a random sample of the data the population mean length is probability not going to be equal. Exercise 2 #A Null:The population mean reaction time between men and women are equal. Alternative: The population mean reaction time between men and women are not equal.

1/23/24, 9:32 PM Math411 Key_Lab 2 file:///Users/zacharykey/Math411-Key_Lab-1.html 3/10 Assumptions: 1.Samples are collected randomly from each of the two populations of interest. 2.The variable is measured on a continuous scale 3.The variable is approximately normally distributed #B mean_1=170.21 mean_2=181.31 s_1=32.643 s_2=45.988 n_1=58 n_2=68 ttest2=(mean_1-mean_2)/sqrt(((s_1*s_1)/n_1)+(s_2*s_2)/n_2) ttest2 ## [1] -1.578112 tcrit2=qt(.05/2,n_1-1,lower.tail = F) tcrit2 ## [1] 2.002465 Taking the absolute value of our t-value it comes out to be 1.58 while the t-critical value is 2.002. Since the t- critical value is greater that means we fail to reject the null. This means that there is a high chance that the population times are equal to each other. #C lower=(mean_1-mean_2)-tcrit*sqrt((s_1^2/n_1)+(s_2^2/n_2)) upper=(mean_1-mean_2)+tcrit*sqrt((s_1^2/n_1)+(s_2^2/n_2)) bin2=c(lower,upper) bin2 ## [1] -25.327044 3.127044 #D In part B the comparison between the t-value and t-critical value supported that the population mean times were equal to each other. In part C we were to find the 95% confidence interval in the difference of the means. If they were equal the values should have been zero in this case it goes through zero as the difference ranged from -25.327 to 3.127 meaning that they could be equal.

1/23/24, 9:32 PM Math411 Key_Lab 2 file:///Users/zacharykey/Math411-Key_Lab-1.html 4/10 #E men=rnorm(n=n_1,mean=mean_1,sd=s_1) women=rnorm(n=n_2,mean=mean_2,sd=s_2) t.test(men,women,varequal=T) ## ## Welch Two Sample t-test ## ## data: men and women ## t = -3.2586, df = 123.96, p-value = 0.001445 ## alternative hypothesis: true difference in means is not equal to 0 ## 95 percent confidence interval: ## -35.00852 -8.55077 ## sample estimates: ## mean of x mean of y ## 165.2034 186.9830 The output of the random sample gave a t-value of -1.9544 and a p-value of 0.05298.The p-value is greater than the critical value of 0.05. This means that there is not enough evidence to reject the null. This shows that even in a random sample of the data the population mean times are probability going to be equal. Exercise 3 #A Null Hypothesis:The mean growth length is equal in treated vs untreated plants Alternative Hypothesis: The mean growth length is not equal in treated vs untreated plants. Assumptions: 1.Samples are collected randomly from each of the two populations of interest. 2.The variable is measured on a continuous scale 3.The variable is approximately normally distributed

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