Question

Suppose you have been hired to study Guinea. You randomly sample 30 Guineans and find that that they make on average $970 and that 23 of them speak French. (Suppose that there is good reason to believe that σ = $75) a) You are asked to determine if Guineans make on average less than $1000.Test the hypothesis at the 5% significance level using the rejection region approach.

b) Determine the p-value for the test in part (a). What can you conclude at the 1% significance level?

c) For the test in part (a), determine the probability of making a type II error if the true mean is: i.) $950, ii.) $1250.

d) For the test in part (a), determine the probability of making a type I error if the true mean is: i.) $1050, ii.) $950.

e) Test the claim at the 0.5% significance level using z as the test statistic.

Answer 1

Given that,
population mean(u)=1000
standard deviation, σ =75
sample mean, x =970
number (n)=30
null, Ho: μ=1000
alternate, H1: μ<1000
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 = 970-1000/(75/sqrt(30)
zo = -2.191
| zo | = 2.191
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.191 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2.191 ) = 0.014
hence value of p0.05 > 0.014, here we reject Ho
ANSWERS
---------------
a.
null, Ho: μ=1000
alternate, H1: μ<1000
test statistic: -2.191
critical value: -1.645
decision: reject Ho
p-value: 0.014
we have enough evidence to support the claim that if Guineans make on average less than $1000.
b.
Given that,
population mean(u)=1000
standard deviation, σ =75
sample mean, x =970
number (n)=30
null, Ho: μ=1000
alternate, H1: μ<1000
level of significance, α = 0.01
from standard normal table,left tailed z α/2 =2.326
since our test is left-tailed
reject Ho, if zo < -2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 970-1000/(75/sqrt(30)
zo = -2.191
| zo | = 2.191
critical value
the value of |z α| at los 1% is 2.326
we got |zo| =2.191 & | z α | = 2.326
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -2.191 ) = 0.014
hence value of p0.01 < 0.014, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=1000
alternate, H1: μ<1000
test statistic: -2.191
critical value: -2.326
decision: do not reject Ho
p-value: 0.014
we do not have enough evidence to support the claim that if Guineans make on average less than $1000.
c.
i.
true mean is $950
Given that,
Standard deviation, σ =75
Sample Mean, X =970
Null, H0: μ=1000
Alternate, H1: μ<1000
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-1000)/75/√(n) < -1.6449 OR if (x-1000)/75/√(n) > 1.6449
Reject Ho if x < 1000-123.3675/√(n) OR if x > 1000-123.3675/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 30 then the critical region
becomes,
Reject Ho if x < 1000-123.3675/√(30) OR if x > 1000+123.3675/√(30)
Reject Ho if x < 977.4763 OR if x > 1022.5237
Implies, don't reject Ho if 977.4763≤ x ≤ 1022.5237
Suppose the true mean is 950
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(977.4763 ≤ x ≤ 1022.5237 | μ1 = 950)
= P(977.4763-950/75/√(30) ≤ x - μ / σ/√n ≤ 1022.5237-950/75/√(30)
= P(2.0066 ≤ Z ≤5.2964 )
= P( Z ≤5.2964) - P( Z ≤2.0066)
= 1 - 0.9776 [ Using Z Table ]
= 0.0224
For n =30 the probability of Type II error is 0.0224
ii.
true mean is 1250$
Given that,
Standard deviation, σ =75
Sample Mean, X =970
Null, H0: μ=1000
Alternate, H1: μ<1000
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.6449
Since our test is left-tailed
Reject Ho, if Zo < -1.6449 OR if Zo > 1.6449
Reject Ho if (x-1000)/75/√(n) < -1.6449 OR if (x-1000)/75/√(n) > 1.6449
Reject Ho if x < 1000-123.3675/√(n) OR if x > 1000-123.3675/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 30 then the critical region
becomes,
Reject Ho if x < 1000-123.3675/√(30) OR if x > 1000+123.3675/√(30)
Reject Ho if x < 977.4763 OR if x > 1022.5237
Implies, don't reject Ho if 977.4763≤ x ≤ 1022.5237
Suppose the true mean is 1250
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(977.4763 ≤ x ≤ 1022.5237 | μ1 = 1250)
= P(977.4763-1250/75/√(30) ≤ x - μ / σ/√n ≤ 1022.5237-1250/75/√(30)
= P(-19.9023 ≤ Z ≤-16.6125 )
= P( Z ≤-16.6125) - P( Z ≤-19.9023)
= 0 - 0 [ Using Z Table ]
= 0
For n =30 the probability of Type II error is 0
d.
i.
truemean is $1050
Given that,
Standard deviation, σ =75
Sample Mean, X =970
Null, H0: μ=1000
Alternate, H1: μ<1000
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-1000)/75/√(n) < -1.645 OR if (x-1000)/75/√(n) > 1.645
Reject Ho if x < 1000-123.375/√(n) OR if x > 1000-123.375/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 30 then the critical region
becomes,
Reject Ho if x < 1000-123.375/√(30) OR if x > 1000+123.375/√(30)
Reject Ho if x < 977.4749 OR if x > 1022.5251
Suppose the true mean is 1050
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(977.4749 < x OR x >1022.5251 | μ1 = 1050)
= P(977.4749-1050/75/√(30) < x - μ / σ/√n OR x - μ / σ/√n >1022.5251-1050/75/√(30)
= P(-5.2965 < Z OR Z >-2.0065 )
= P( Z <-5.2965) + P( Z > -2.0065)
= 0 + 0.9776 [ Using Z Table ]
= 0.9776
ii.
true mean is $950
Given that,
Standard deviation, σ =75
Sample Mean, X =970
Null, H0: μ=1000
Alternate, H1: μ<1000
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-1000)/75/√(n) < -1.645 OR if (x-1000)/75/√(n) > 1.645
Reject Ho if x < 1000-123.375/√(n) OR if x > 1000-123.375/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 30 then the critical region
becomes,
Reject Ho if x < 1000-123.375/√(30) OR if x > 1000+123.375/√(30)
Reject Ho if x < 977.4749 OR if x > 1022.5251
Suppose the true mean is 950
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(977.4749 < x OR x >1022.5251 | μ1 = 950)
= P(977.4749-950/75/√(30) < x - μ / σ/√n OR x - μ / σ/√n >1022.5251-950/75/√(30)
= P(2.0065 < Z OR Z >5.2965 )
= P( Z <2.0065) + P( Z > 5.2965)
= 0.9776 + 0 [ Using Z Table ]
= 0.9776
e.
Given that,
population mean(u)=1000
standard deviation, σ =75
sample mean, x =970
number (n)=30
null, Ho: μ=1000
alternate, H1: μ<1000
level of significance, α = 0.005
from standard normal table,left tailed z α/2 =2.576
since our test is left-tailed
reject Ho, if zo < -2.576
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 970-1000/(75/sqrt(30)
zo = -2.191
| zo | = 2.191
critical value
the value of |z α| at los 0.5% is 2.576
we got |zo| =2.191 & | z α | = 2.576
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : left tail - ha : ( p < -2.191 ) = 0.014
hence value of p0.005 < 0.014, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=1000
alternate, H1: μ<1000
test statistic: -2.191
critical value: -2.576
decision: do not reject Ho
p-value: 0.014
we do not have enough evidence to support the claim that if Guineans make on average less than $1000.