Question

In: Statistics and Probability

A hypothesis test was conducted to compare the mean SAT-Verbal score of a sample of 10...

A hypothesis test was conducted to compare the mean SAT-Verbal score of a sample of 10 students from one school to the known national norms. In the sample of 10 students, the mean was 555. In the population, SAT-Verbal scores are normed to have a mean of 500 and standard deviation of 100. The two-tailed single sample mean z test resulted in a test statistic of 1.739 and p value of 0.082. At the 0.05 alpha level, the results were not statistically significant.

A. There are five ways to increase statistical power. In this specific scenario, what are three changesthe researchers could realistically make to increase statistical power.

B. Of the three changes that the researchers could make, which one would you recommend? Explain why you think this is the best option.

Solutions

Expert Solution

Given that,
population mean(u)=500
standard deviation, σ =100
sample mean, x =555
number (n)=10
null, Ho: μ=500
alternate, H1: μ!=500
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 555-500/(100/sqrt(10)
zo = 1.739
| zo | = 1.739
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.739 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.739 ) = 0.082
hence value of p0.05 < 0.082, here we do not reject Ho
ANSWERS
---------------
B.
null, Ho: μ=500
alternate, H1: μ!=500
test statistic: 1.739
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.082
we do not have enough evidnce to supportthe claim that SAT-Verbal scores are normed to have a mean of 500.
A.
Given that,
Standard deviation, σ =100
Sample Mean, X =555
Null, H0: μ=500
Alternate, H1: μ!=500
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-500)/100/√(n) < -1.96 OR if (x-500)/100/√(n) > 1.96
Reject Ho if x < 500-196/√(n) OR if x > 500-196/√(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 500-196/√(10) OR if x > 500+196/√(10)
Reject Ho if x < 438.0194 OR if x > 561.9806
Implies, don't reject Ho if 438.0194≤ x ≤ 561.9806
Suppose the true mean is 555
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(438.0194 ≤ x ≤ 561.9806 | μ1 = 555)
= P(438.0194-555/100/√(10) ≤ x - μ / σ/√n ≤ 561.9806-555/100/√(10)
= P(-3.6993 ≤ Z ≤0.2207 )
= P( Z ≤0.2207) - P( Z ≤-3.6993)
= 0.5873 - 0.0001 [ Using Z Table ]
= 0.5872
For n =10 the probability of Type II error is 0.5872
power = 1-type 2 error
power = 1-0.5872
power = 0.4128


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