Solutions of QUIZ 5. M262 Section 1. (25.04.2022).
1. A zoologist is going to test a Null Hypothesis that the standard deviation σ
of the weights of frogs in some region is equal to 5 grams. She collected 11 frogs,
measured their weights and found a sample standard deviation s = 3. Based on this
data should Null Hypothesis σ = 5 be rejected in favor of Alternative Hypothesis
σ 6= 5? Use α = 0.1.
Solution:
H0 : σ = 5
H1 : p 6= 5
For d = 10 χ2 (0.95) = 3.94 and χ2 (0.95) = 18.31. Therefore, the rejection region is
[0, 3.94] ∪ [18.31, ∞). Since
s2 9
(n − 1) · = 10 · = 3.6 < 3.94
σ2 25
we reject H0 and accept H1 .
2. We compare the number of words per sentence in two magazines. On the
basis of 50 randomly selected sentences from the first and 60 randomly selected
sentences from the second magazine we find x̄ = 12.7, s1 = 4 and ȳ = 15.5, s2 = 5,
respectively. Determine a 90% confidence interval for the difference in mean number
of words per sentence.
Solution:
From z-table we get zα/2 = 1.65. Therefore, a 90 % confidence interval is
s s
42 52 42 52
(12.7 − 15.5 − 1.65 · + , 12.7 − 15.5 + 1.65 · + )
50 60 50 60
= (−2.8 − 1.42, −2.8 + 1.42) = (−4.22, −1.38)
1
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:07:41 GMT -05:00
https://www.coursehero.com/file/249824299/22262q5solpdf/
1. A zoologist is going to test a Null Hypothesis that the standard deviation σ
of the weights of frogs in some region is equal to 5 grams. She collected 11 frogs,
measured their weights and found a sample standard deviation s = 3. Based on this
data should Null Hypothesis σ = 5 be rejected in favor of Alternative Hypothesis
σ 6= 5? Use α = 0.1.
Solution:
H0 : σ = 5
H1 : p 6= 5
For d = 10 χ2 (0.95) = 3.94 and χ2 (0.95) = 18.31. Therefore, the rejection region is
[0, 3.94] ∪ [18.31, ∞). Since
s2 9
(n − 1) · = 10 · = 3.6 < 3.94
σ2 25
we reject H0 and accept H1 .
2. We compare the number of words per sentence in two magazines. On the
basis of 50 randomly selected sentences from the first and 60 randomly selected
sentences from the second magazine we find x̄ = 12.7, s1 = 4 and ȳ = 15.5, s2 = 5,
respectively. Determine a 90% confidence interval for the difference in mean number
of words per sentence.
Solution:
From z-table we get zα/2 = 1.65. Therefore, a 90 % confidence interval is
s s
42 52 42 52
(12.7 − 15.5 − 1.65 · + , 12.7 − 15.5 + 1.65 · + )
50 60 50 60
= (−2.8 − 1.42, −2.8 + 1.42) = (−4.22, −1.38)
1
This study source was downloaded by 100000899606396 from CourseHero.com on 09-25-2025 13:07:41 GMT -05:00
https://www.coursehero.com/file/249824299/22262q5solpdf/
