Question

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 .

Answer 1

Answer with explanation:

Test hypothesis :

$H_0 : \mu =5\\\\ H_a: \mu >5$

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

Test statistic : $z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

, where $\overline{x}$ = sample mean

$\mu$ = population mean

$\sigma$ =population standard deviation

n= Sample size

Substitute values, we get

$z=\dfrac{ 5.3-5}{\dfrac{1}{\sqrt{500}}}$

$z=\dfrac{ 0.3}{0.04472135955}\approx6.7$

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. $H_a$ is accepted.

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