Question

Past test scores in my math course typically have a standard deviation equal to 14.1. The class from last semester had 27 test scores with a standard deviation of 9.3. Use a 0.01 significance level to test the claim that the class from last semester has less variation than the other classes from the past. Assume a simple random sample is selected from a normally distributed population.

1. Write the conclusion utilizing the “correct” words using Table 8-3, p. 366. Make sure you insert the relevant portions of the claim into the conclusion.
2. Construct the associated confidence interval (CI).
3. State the CI using the correct notation using the correct notation.
4. If applicable, state the margin of error (E = __) and show your calculations.
5. If applicable, state the point estimate using the correction notation and show your calculations.
6. Does the CI support the hypothesis test conclusion? Explain/interpret. This may require a sentence or two, not just a single word.
7. Express the Type I error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).
8. Express the Type II error in the context of the problem using the words from the “helper” document. Make sure the conclusion is worded such that it addresses the claim (p. 368).

Answer 1

Given that,
population standard deviation (σ)=9.3
sample standard deviation (s) =14.1
sample size (n) = 27
we calculate,
population variance (σ^2) =86.49
sample variance (s^2)=198.81
null, Ho: σ =9.3
alternate, H1 : σ <9.3
level of significance, α = 0.01
from standard normal table,left tailed ᴪ^2 α/2 =45.642
since our test is left-tailed
reject Ho, if ᴪ^2 o < -45.642
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(27 - 1 ) * 198.81 / 86.49 = 26*198.81/86.49 = 59.765
| ᴪ^2 cal | =59.765
critical value
the value of |ᴪ^2 α| at los 0.01 with d.f (n-1)=26 is 45.642
we got | ᴪ^2| =59.765 & | ᴪ^2 α | =45.642
make decision
hence value of | ᴪ^2 cal | > | ᴪ^2 α| and here we reject Ho
ᴪ^2 p_value =0.0002
ANSWERS
---------------
a.
null, Ho: σ =9.3
alternate, H1 : σ <9.3
test statistic: 59.765
critical value: -45.642
p-value:0.0002
decision: reject Ho
we have enough evidence to support the claim that the class from last semester has less variation than the other classes from the past.
b.
A confidence interval calculates the probability that a population parameter will fall between two set values. Confidence intervals measure the degree of uncertainty or certainty in a sampling method.
Most often, confidence intervals reflect confidence levels of 95% or 99%.
c.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.01
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.99)/2 = 0.01/2 = 0.005
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.005 = 0.995
the two critical values ᴪ^2 left, ᴪ^2 right at 26 df are 48.2899 , 11.16
s.d( s )=9.3
sample size(n)=27
confidence interval for σ^2= [ 26 * 86.49/48.2899 < σ^2 < 26 * 86.49/11.16 ]
= [ 2248.74/48.2899 < σ^2 < 2248.74/11.1602 ]
[ 46.5675 < σ^2 < 201.4964 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (46.5675) < σ < sqrt(201.4964), ]
= [ 6.824 < σ < 14.1949 ]
d.
not possible margin of error
e.
yes,
the CI support the hypothesis test conclusion
we have enough evidence to support the claim that the class from last semester has less variation than the other classes from the past.
f.
type 1 error is possible because it reject thenull hypothesis.
g.
type 2 error is possible only when its fails to reject the null hypothesis.