The community bank survey asked about net income and reported the percent change in net income...

The community bank survey asked about net income and reported the percent change in net income between the first half of last year and the first half of this year. The mean change for the 120 banks in the sample is 7.6%. Because the sample size is large, we are willing to use the sample standard deviation s = 20.5% as if it were the population standard deviation σ. Does the 7.6% mean increase provide evidence that the net income for all banks has changed? Assumeα= 0.05. You must show all steps as outlined in class to receive full credit.

Solutions

Expert Solution

Given that,
population mean(u)=7.6
standard deviation, sigma =20.5
sample mean, x =7.6
number (n)=120
null, Ho: μ=7.6
alternate, H1: μ>7.6
level of significance, alpha = 0.05
from standard normal table,right tailed z alpha/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 7.6-7.6/(20.5/sqrt(120)
zo = 0
| zo | = 0
critical value
the value of |z alpha| at los 5% is 1.645
we got |zo| =0 & | z alpha | = 1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : right tail - ha : ( p > 0 ) = 0.5
hence value of p0.05 < 0.5, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=7.6
alternate, H1: μ>7.6
test statistic: 0
critical value: 1.645
decision: do not reject Ho
p-value: 0.5
we do not have enough evidence to support the claim that mean increase provide evidence that the net income for all banks has changed

orchestra answered 1 year ago

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