Question

An analyst on Fox News unequivocally claims that the average years of education obtained by an adult Mexican immigrant in the U.S. is at most 8 years. On the other hand, an average native receives 13 years or more of education. This, points the analyst, is an important reason why wages of Mexican immigrants lag behind natives.

a) The analyst backs his claim saying he took a sample of 100 Mexican immigrants and found that their average education = 8.7 years. The standard deviation in education of the population of Mexican immigrants is 4.5 years. He conducted his test at 5%. You take a look at his test set-up. You see his null hypothesis was ”Mexican immigrants at most get 8 years of education”. You use the same sample as he does; you have access to the same population data as he does. You also conduct the test at 5% significance level. Can you agree with the analyst’s conclusion? Why or why not?

b) Find the p-value of the test in part (a). What is the probability of Type I error in part (a)? If the analyst added another 200 immigrants to his sample (everything else constant) how would it affect the value of Type I error?

Answer 1

Given that,
population mean(u)=8
standard deviation, σ =4.5
sample mean, x =8.7
number (n)=100
null, Ho: μ=8
alternate, H1: μ<=8
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 8.7-8/(4.5/sqrt(100)
zo = 1.5556
| zo | = 1.5556
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.5556 & | z α | = 1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < 1.5556 ) = 0.9401
hence value of p0.05 < 0.9401, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: μ=8
alternate, H1: μ<=8
test statistic: 1.5556
critical value: -1.645
decision: do not reject Ho
b.
p-value: 0.9401
we do not have enough evidence to support the claim that the average years of education obtained by an adult
Mexican immigrant in the U.S. is at most 8 years.
type 1 error is possible when reject the null hypothesis.

c.
Given that,
Standard deviation, σ =4.5
Sample Mean, X =8.7
Null, H0: μ=8
Alternate, H1: μ<=8
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.645
Since our test is left-tailed
Reject Ho, if Zo < -1.645 OR if Zo > 1.645
Reject Ho if (x-8)/4.5/√(n) < -1.645 OR if (x-8)/4.5/√(n) > 1.645
Reject Ho if x < 8-7.4025/√(n) OR if x > 8-7.4025/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 100 then the critical region
becomes,
Reject Ho if x < 8-7.4025/√(100) OR if x > 8+7.4025/√(100)
Reject Ho if x < 7.2598 OR if x > 8.7403
Suppose the true mean is 8.7
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(7.2598 < x OR x >8.7403 | μ1 = 8.7)
= P(7.2598-8.7/4.5/√(100) < x - μ / σ/√n OR x - μ / σ/√n >8.7403-8.7/4.5/√(100)
= P(-3.2004 < Z OR Z >0.0896 )
= P( Z <-3.2004) + P( Z > 0.0896)
= 0.0007 + 0.4643 [ Using Z Table ]
= 0.465

d.
If the analyst added another 200 immigrants to his sample.
yes,
it affect the value of Type I error
Given that,
Standard deviation, σ =4.5
Sample Mean, X =8.7
Null, H0: μ=8
Alternate, H1: μ<=8
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.645
Since our test is left-tailed
Reject Ho, if Zo < -1.645 OR if Zo > 1.645
Reject Ho if (x-8)/4.5/√(n) < -1.645 OR if (x-8)/4.5/√(n) > 1.645
Reject Ho if x < 8-7.4025/√(n) OR if x > 8-7.4025/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 200 then the critical region
becomes,
Reject Ho if x < 8-7.4025/√(200) OR if x > 8+7.4025/√(200)
Reject Ho if x < 7.4766 OR if x > 8.5234
Suppose the true mean is 8.7
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(7.4766 < x OR x >8.5234 | μ1 = 8.7)
= P(7.4766-8.7/4.5/√(200) < x - μ / σ/√n OR x - μ / σ/√n >8.5234-8.7/4.5/√(200)
= P(-3.8448 < Z OR Z >-0.555 )
= P( Z <-3.8448) + P( Z > -0.555)
= 0.0001 + 0.7106 [ Using Z Table ]
= 0.7107