Estimating Single Variance or Stand Dev Worksheet Answers
1. Let 𝑥 = age in years of a rural Quebec woman at the time of her first marriage. IN
the year 1941, the population variance of 𝑥 was approximately 𝜎 2 = 5.1. Suppose a
recent study of age at first marriage for a random sample of 41 women in rural
Quebec gave a sample variance of 𝑠 2 = 3.3. Use a 5% level of significance to test the
claim that the current variance is less than 5.1. Find a 90% confidence interval for
the population variance and population standard deviation.
𝐻0 ∶ 𝜎 2 = 5.1
𝑥2 =
(𝑛−1)𝑠 2
𝜎2
𝐻1 ∶ 𝜎 2 < 5.1
=
(41−1)(3.3)
(5.1)
=
(40)(3.3)
5.1
𝛼 = 0.05
= 25.89
𝑑. 𝑓. = 𝑛 − 1 = 41 − 1 = 40
𝑃 − 𝑣𝑎𝑙𝑢𝑒 ( left – tailed ) between 0.975 and 0.950
1 − 0.975 < 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 1 − 0.950
0.025 < 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.050
Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.05, reject null
Confidence interval ( 𝜎 2 )
𝑥𝑈2 =
1−𝑐
𝑥𝐿2 =
1+𝑐
2
2
(𝑛−1)𝑠 2
2
𝑥𝑈
(40)(3.3)
55.76
=
=
1−0.90
2
1+0.90
2
< 𝜎2 <
< 𝜎2 <
=
=
0.10
2
1.90
2
= 0.05 with d.f. = 40 𝑥𝑈2 = 55.76 ( from table )
= 0.95 with d.f. = 40 𝑥𝐿2 = 26.51 ( from table )
(𝑛−1)𝑠 2
𝑥𝐿2
(40)(3.3)
26.51
Confidence interval (𝜎)
√2.37 < 𝜎 < √5.00
1.54 < 𝜎 < 2.24
2.37 < 𝜎 2 < 5.00
2. Let 𝑥 represent the number of mountain climbers killed each year. The long – term
variance of 𝑥 is approximately 𝜎 2 = 136.2. Suppose that for the past 8 years, the
variance has been 𝑠 2 = 115.1 . Use a 1% level of significance to test the claim that
the recent variance for number of mountain climber deaths is less than 136.2. Find a
90% confidence interval for the population variance and population standard
deviation.
𝐻0 ∶ 𝜎 2 = 136.2
𝑥2 =
(𝑛−1)𝑠 2
(8−1)(115.1)
=
𝜎2
𝐻1 ∶ 𝜎 2 < 136.2
136.2
=
(7)(115.1)
136.2
𝛼 = 0.01
= 5.92
𝑑. 𝑓. = 𝑛 − 1 = 8 − 1 = 7
𝑃 − 𝑣𝑎𝑙𝑢𝑒 ( left – tailed ) between 0.100 and 0.900
1 − 0.900 < 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 1 − 0.100
0.100 < 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.900
Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject null hypothesis
Confidence interval ( 𝜎 2 )
𝑥𝑈2 =
1−𝑐
𝑥𝐿2 =
1+𝑐
2
(𝑛−1)𝑠 2
2
𝑥𝑈
2
=
1−0.90
2
1+0.90
2
< 𝜎2 <
(7)(115.1)
14.07
=
0.10
=
=
2
1.90
2
= 0.05 with d.f. = 7 𝑥𝑈2 = 14.07 ( from table )
= 0.95 with d.f. = 7 𝑥𝐿2 = 2.17 ( from table )
(𝑛−1)𝑠 2
< 𝜎2 <
𝑥𝐿2
(7)(115.1)
2.17
Confidence interval (𝜎)
√57.26 < 𝜎 < √371.29
7.57 < 𝜎 < 19.27
57.26 < 𝜎 2 < 371.29
3. A set of solar batteries is used in a research satellite. The satellite can run on only
one battery, but it runs best if more than one battery is used. The variance 𝜎 2 of
lifetimes of these batteries affects the useful lifetime of the satellite before it goes
dead. Engineers have determined that a variance of 𝜎 2 = 23 months is most
desirable for these batteries. A random sample of 22 batteries gave a sample
variance of 14.3 months. Using a 0.05 level of significance, test the claim that 𝜎 2 =
23 against the claim that 𝜎 2 is different from 23. Find a 90% confidence interval for
both the population variance and population standard deviation.
𝐻0 : 𝜎 2 = 23
𝑥2 =
(𝑛−1)𝑠 2
𝜎2
𝐻1 ∶ 𝜎 2 ≠ 23
=
(22−1)(14.3)
23
=
(21)(14.3)
23
𝛼 = 0.05
= 13.06
𝑑. 𝑓. = 𝑛 − 1 = 22 − 1 = 21
𝑃 − 𝑣𝑎𝑙𝑢𝑒 is between 0.950 and 0.900 ( two – tailed )
1 − 0.950 <
0.050 <
𝑃−𝑣𝑎𝑙𝑢𝑒
2
𝑃−𝑣𝑎𝑙𝑢𝑒
2
< 1 − 0.900
< 0.100
0.100 < 𝑃 − 𝑣𝑎𝑙𝑢𝑒 < 0.200
Conclude the test
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject null
Confidence interval ( 𝜎 2 )
𝑥𝑈2 =
1−𝑐
𝑥𝐿2 =
1+𝑐
2
2
(𝑛−1)𝑠 2
2
𝑥𝑈
=
1−0.90
2
1+0.90
2
< 𝜎2 <
(21)(14.3)
32.67
=
=
=
0.10
2
1.90
2
= 0.05 with d.f. = 21 𝑥𝑈2 = 32.67 ( from table )
= 0.95 with d.f. = 21 𝑥𝐿2 = 11.59 ( from table )
(𝑛−1)𝑠 2
< 𝜎2 <
𝑥𝐿2
(21)(14.3)
11.59
Confidence interval (𝜎)
√9.19 < 𝜎 < √25.91
3.03 < 𝜎 < 5.09
9.19 < 𝜎 2 < 25.91