Question

Assume Y is a normally distributed population with unknown mean, μ, and σ = 25. If we desire to test  H o: μ = 200 against H 1: μ < 200,

(a.) What test statistic would you use for this test?

(b.)What is the sampling distribution of the test statistic?

(c.) Suppose we use a critical value of 195 for the scenario described. What is the level of significance of the resulting test?

(d.) Staying with a critical value of 195, find the probability of Type II error if in fact μ = 198.

(e.) Staying with a critical value of 195, state the point on the power curve corresponding to your answer to part (d.)

(f.) For any test of hypothesis, we know a point on the power curve automatically without doing any Z-value calculations. What is this point for the test described in part (c.)?

Answer 1

ANSWER:

Given that,
population mean(u) =200
standard deviation,

σ =25
sample mean,

x =198
number (n)=196
null, Ho: μ=200
alternate, H1: μ<200
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 = 198-200/(25/sqrt(196)
zo = -1.12
| zo | = 1.12
critical value
the value of |z α|

at los 5% is 1.645
we got |zo| =1.12 &

| 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 < -1.12 ) = 0.131
hence value of p0.05 < 0.131, here we do not reject Ho
ANSWERS
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(a)
null, Ho: μ=200
alternate, H1: μ<200
(b)
test statistic: -1.12
(c)
critical value: -1.645
decision: do not reject Ho
p-value: 0.131
we do not have enough evidence to support the claim that mean is less than 200.
(d)
Given that,
Standard deviation, σ =25
Sample Mean, X =198
Null, H0: μ=200
Alternate, H1: μ<200
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-200)/25/√(n) < -1.6449 OR

if (x-200)/25/√(n) > 1.6449
Reject Ho if x < 200-41.1225/√(n) OR

if x > 200-41.1225/√(n)
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Suppose the size of the sample is n = 196 then the critical region
becomes,
Reject Ho if x < 200-41.1225/√(196) OR if x > 200+41.1225/√(196)
Reject Ho if x < 197.0627 OR if x > 202.9373
Implies, don't reject Ho if 197.0627≤ x ≤ 202.9373
Suppose the true mean is 198
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(197.0627 ≤ x ≤ 202.9373 | μ1 = 198)
= P(197.0627-198/25/√(196) ≤ x - μ / σ/√n ≤ 202.9373-198/25/√(196)
= P(-0.5249 ≤ Z ≤2.7649 )
= P( Z ≤2.7649) - P( Z ≤-0.5249)
= 0.9972 - 0.2998 [ Using Z Table ]
= 0.6974
For n =196 the probability of Type II error is 0.6974
power of the test = 1-type 2 error
(f)
power of the test = 1-0.6974
power = 0.3026