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.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2