Suppose the time that it takes a certain large bank to approve a home loan is normally distributed, with mean (in days) μ μ and standard deviation σ = 1 σ=1 . the bank advertises that it approves loans in 5 days, on average, but measurements on a random sample of 500 loan applications to this bank gave a mean approval time of ¯ x = 5.3 x¯=5.3 days. is this evidence that the mean time to approval is actually longer than advertised? to answer this, test the hypotheses h 0 : μ = 5 h0: μ=5 , h α : μ > 5 hα: μ> 5 at significance level α = 0.01 α=0.01 .

Test hypothesis :

Since alternative hypothesis is right-tailed and population standard deviation is known σ = 1 , so we perform a right-tailed  z-test.

Test statistic :

, where = sample mean

= population mean

=population standard deviation

n= Sample size

Substitute values, we get

Critical value for 0.01 significance level in z-table is 2.326.

Decision : Test statistic (6.7)> Critical value ( 2.326), it means we reject that null hypothesis.

i.e. is accepted.

We conclude that there is sufficient evidence that the mean time to approval is actually longer than advertised.

idk

step-by-step explanation:

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.