I need help with this as I have to round the decimal answers to at least 3 places, unless otherwise, but I am getting different numbers every time when I use the code and I wanted to see If I can match with someone to see if I am correct. I am using the R coding program to solve the question and I have also included below snips from a review sheet I did with the code.

The question:

7. Do forest fires near streams hurt fishing in those streams in future years. There are arguments for and against thesis thesis. The mean number of trout, Ux caught by an angler in a day of fishing in the Yellowstone River before the 1988 super fires will be compared to the mean number of trout, My caught by an angler in the same river in 1993. The true values of ux and My are unknown. It is recognized that the true standard deviations are Ox = 1.9 for the before 1988 measurements and oy = 2.3 for the 1993 measurements. We take a random sample of m = 167 angler catches before 1988 and a random sample of n=125 angler catches in 1993. The mean angler catches were were x = 5.2 for the before 1988 measurements and y = 6.8 for the 1993 measurements. Assuming independence between the years and assuming angler catch amounts are normally distributed we would like to estimate Ux - My. a. What is the standard deviation of the distribution of x ? b. What is the standard deviation of the distribution of x - y? c. Create a 94% confidence interval for Ux - My? d. What is the length of the confidence interval in part c? e. If we let n stay 125 but vary m, what is the smallest m for which the length of the 96% confidence interval would be 0.88 or less? > #7 trout catch before and after forest fires >m ybar sigmax sigmay zstar zstar [1] 1.880794 > # 94 % confidence interval > c(xbar - ybar - zstar*sediff, xbar - ybar + zstar*sediff) #c [1] -2. 075572 -1. 124428 >#d Length of conf int: > 2*zstar*sediff #d length of conf int. [1] 0.9511444 >#e) Let m vary. Smallest m for which length of confidence interval is . 88 or less. fl ll LL [1] 290.9014 ceiling(11) # smallest m [1] 291Who studies more, college freshmen or sophomores? The mean number of hours, My , that freshmen study per week will be compared to the mean number of hours, My that sophomores study per week. The true values of My and My are unknown. It is recognized that the true standard deviations are ox = 6.3 hours for freshmen and oy = 5.8 hours for sophomores. We take a random sample of m = 27 freshmen and n= 38 sophomores. The mean numbers of hours studying were were x= 39.6 for the freshmen and y= 35.4 for the sophomores. Assuming all measurements were independent from one another and assuming that hours studied are normally distributed we would like to estimate Mx - My . a) What is the standard deviation of the distribution of y? b) What is the standard deviation of the distribution of x - y ? c) Create a 98% confidence interval for MX - My ? d) What is the length of the confidence interval in part c) ? e) If we let n stay at 38 but vary m, what is the smallest m for which the length of the 98% confidence interval would be 5.0 or less