Question

According to the 2010 US Census, the average number of residents per housing

unit for the n=87 counties in Minnesota was 2.10, and the standard deviation

was 0.38. Test whether the true mean number of residents per housing unit

in Minnesota in 2010 is less than the national value of 2.34 at the level

α

= 0

.

05.

a. Show all five steps of this test.

b. What type of error could we be making in this context?

c. What is the minimum average number of residents per household needed

in order to fail to reject H0? Assume the sample standard deviation is the same.

d. Suppose the true number of residents per household in Minnesota is normally

distributed with a mean of 2.0 and a standard deviation of 0.4. Suppose we reject

the null hypothesis if the sample means the number of residents is less than 2.27. What

is the probability of making a type II error?

Answer 1

a)

Claim : Mean number of residents per housing unit in Minnesota in 2010 is less than the national value of 2.34

H₀ : µ = 2.34 vs H_a : µ < 2.34

Given : = 2.10 , s = 0.38 , n = 87

Population standard deviation σ is unknown therefore we use t statistic.

Test statistic:

t = =  

t = -5.87

Critical value:     α = 0.05

As H_a contain < sign , this is left tail test,critical value would be negative.

df =n-1 = 87 - 1 = 86

Therefore t_{(0.05,86 )} = -1.663

Decision: As t is less than -1.663 , we reject H₀

Conclusion: There is significant evidence that mean number of residents per housing unit in Minnesota in 2010 is less than the national value of 2.34

b) Type I error : We reject H₀, when it is true .

we could be making Type I error in this context

c) -1.663 = =

= (-1.663 * 0.0407) + 2.34

= 2.27

d) P( Type II error ) = P( Fail to reject H₀ Given µ = 2 and σ = 0.4 )

= P( > 2.27 )

=

= P( z > 6.30 )

=1 - P( z  ≤ 6.30 )

= 1 - 1

= 0

Probability of making a type II error is 0