Question: i need help with my homework. need it ASAP. The weights of a random sample of 15 marathon runners were taken before and after running

i need help with my homework. need it ASAP.

a random sample of 15 marathon runners were taken before and after

26 miles that takes more than 2 hours for elite athletes. Please

The weights of a random sample of 15 marathon runners were taken before and after running a marathon race, a long distance race of about 42km or 26 miles that takes more than 2 hours for elite athletes. Please run the following lines of code in R to save the before and after values: before t*), we REJECT the null hypothesis at the a = 0.05 level. There IS statistically significant evidence that the weight of marathon runners changes after a running a marathon race. Since the test statistic IS more extreme than the critical value (t > t*), we FAIL TO REJECT the null hypothesis at the a = 0.05 level. There IS NOT statistically significant evidence that the weight of marathon runners changes after a running a marathon race. O Since the test statistic IS NOT more extreme than the critical value (t 0 O Ho : Xfert. - Kno fert. = 0 Ha : Xfert. - no fert. > 0 O Ho : Mfert. - Mno fert. = 0 Ha : Mfert. - Mno fert. 0A farmer wants to see if a new kind of fertilizer can help him grow better fruits. He prepares two separate plots of land and adds the fertilizer in one of those only. After a month he measure the weights of the fruits in each plot. Here are the results: plot 1 (no fertilizer) plot2 (fertilizer) sample standard deviation 3m fart\" = 0.8 Sfert, = 0.9 Should this farmer invest in fertilizers? That is, is the mean weight of fruits signicantly higher for the plot with fertilizer compared to the plot without it at the a = 0,1 level. Compute the test statistic for this test. 5 3.83.5 O z = ':\" \"\"8\" = 2 2 = 2.461955 dfert trio fert %+% \"fart \"no ferh :E 5 3.83.5 13 = fe't \"\"6\" = = 2.662387 83.5 .30 its E+E nfert \"no ferl'. 110 120 5 5 3.83.5 O t = :tf\" = T = 2.461955 sfe; s11.0 felt + \"felt \"no fart 110 120 5 ,5 3.83.5 O z: % = = 2.662387 dfert orig ferl: + 110 120 \"fart \"no fert

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