Assume that both populations are normally distributed. (a) Test whether mu 1 not equals mu 2 at the alpha equals 0.01 level of significance for the given sample data. (b) Construct a 99% confidence interval about mu 1 minus mu 2. Population 1 Population 2 n 10 10 x overbar 13.2 11 s 3.7 3.9 (a) Test whether mu 1 not equals mu 2 at the alpha equals 0.01 level of significance for the given sample data. Determine the null and alternative hypothesis for this test.

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol

step-by-step explanation: