Testing Mean and Proportion Differences Worksheet Answers
1. Two populations have mound – shaped, symmetric distributions. A random sample
of 16 measurements from the first population had a sample mean of 20, with sample
standard deviation 2. An independent random sample of 9 measurements from the
second population had a sample mean of 19, with a sample standard deviation of 3.
Test the claim that the population mean of the first population exceeds that of the
second. Use 𝛼 = 0.05
𝐻0 ∶ 𝜇1 = 𝜇2
𝑧=
𝑥̅1 −𝑥̅2
𝜎2 𝜎2
√ 1+ 1
𝑛1 𝑛2
𝐻1 ∶ 𝜇1 > 𝜇2
=
20−19
=
2
2
√2 +3
20 19
𝛼 = 0.05
1
√
=
0.2+0.474
1
√0.674
= 1.218
𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 1.218) = 1 − 0.8997 = 0.1003
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject 𝐻0
2. A random sample of 49 measurements from a population with a population
standard deviation 3 had a sample mean of 10. An independent random sample of
64 measurements from a second population with population standard deviation 4
had a sample mean of 12. Test the claim that the population means are different. Use
a 1% level of significance
𝐻0 ∶ 𝜇1 = 𝜇2
𝑧=
𝑥̅ 1 −𝑥̅ 2
𝜎2 𝜎2
√ 1+ 1
𝑛1 𝑛2
=
𝐻1 ∶ 𝜇1 ≠ 𝜇2
20−19
2
2
√2 +3
20 19
=
1
√0.2+0.474
𝛼 = 0.01
=
1
√0.674
= 1.218
** two tailed
𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 1.218) = 2(1 − 0.8997) = 2(0.1003) = 0.2006
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject 𝐻0
3. For one binomial experiment, 75 binomial trials produced 45 successes. For a
second independent binomial experiment, 100 binomial trials produced 70
successes. At the 5% level of significance, test the claim that the probability of
success for the two binomial experiments differ.
𝑟1 = 45 , 𝑛1 = 75 , 𝑟2 = 70 , 𝑛2 = 100
𝐻0 ∶ 𝑝1 − 𝑝2 = 0
𝑝̅ =
𝑟1 +𝑟2
𝑛1 +𝑛2
𝑝̂1 =
𝑧=
𝑟1
𝑛1
=
𝐻1 ∶ 𝑝1 ≠ 𝑝2
45+70
=
75+100
45
=
115
175
= 0.60
75
̅𝑞
̅
𝑝
=
̅𝑞
̅
𝑝
√𝑛 +𝑛
1
2
= 0.657
𝑝̂2 =
𝑝̂1 −𝑝̂2
𝛼 = 0.05
𝑟2
𝑛2
=
𝑞̅ = 1 − 𝑝̅ = 1 − 0.657 = 0.343
70
100
= 0.70
−0.10
√
(0.657)(0.343) (0.657)(0.343)
+
75
100
=
𝑝̂1 − 𝑝̂2 = 0.60 − 0.70 = −0.10
−0.10
√0.003+0.002
=
−0.10
√0.005
= −1.414
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.414) = 2(0.0793) = 0.1586
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we fail to reject 𝐻0
4. REM ( rapid eye movement ) sleep is sleep during which most dreams occur. It is
thought that children have more REM sleep than adults. Assume REM sleep time is
normally distributed for both children and adults.
Children :
𝑛1 = 10 , 𝑥̅1 = 2.8 hours , 𝜎 = 0.5 hours
Adults :
𝑛2 = 10 , 𝑥̅2 = 2.1 hours , 𝜎 = 0.7 hours
Do these data indicate that, on average, children tend to have more REM sleep time
than adults ? Use 𝛼 = 0.01
𝐻0 ∶ 𝜇1 = 𝜇2
𝑧=
𝑥̅1 −𝑥̅2
𝜎2 𝜎2
√ 1+ 1
𝑛1 𝑛2
𝐻1 ∶ 𝜇1 > 𝜇2
=
2.8−2.1
=
2
2
√0.5 +0.7
10
10
𝛼 = 0.01
0.7
√
=
0.025+0.049
𝑃 − 𝑣𝑎𝑙𝑢𝑒(𝑧 > 2.56) = 1 − 0.9948 = 0.0052
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 ≤ 0.01, we reject 𝐻0
0.7
√0.074
= 2.56
5. A random sample of 𝑛1 = 10 regions in New England gave the following violent
crime rates ( per million population )
𝒙𝟏 ∶ New England Crime Rate
3.5
3.7
4.0
3.9
3.3
4.1
1.8
4.8
2.9
3.1
Another random sample of 𝑛2 = 12 regions in the Rocky Mountain states gave the
following violent crime rates ( per million population )
𝒙𝟐 ∶ Rocky Mountain States
3.7
4.3
4.5
5.3
3.3
4.8
3.5
2.4
3.1
3.5
5.2
2.8
Do the data indicate that the violent crime rate in the Rocky Mountain region is
higher than that in New England? Use 𝛼 = 0.01
Using a calculator
𝑥̅1 = 3.51 , 𝑠1 = 0.77 , 𝑛1 = 10
𝑥̅2 = 3.87 , 𝑠2 = 0.94 , 𝑛2 = 12
𝐻0 ∶ 𝜇1 = 𝜇2
𝑡=
𝑥̅1 −𝑥̅ 2
𝑠2 𝑠2
√ 1+ 1
𝑛1 𝑛2
𝐻1 ∶ 𝜇1 < 𝜇2
=
3.51−3.87
=
2
2
√0.77 +0.94
10
12
𝛼 = 0.01
−0.36
√
=
0.059+0.074
−0.36
√0.133
= −0.99
𝑃 − 𝑣𝑎𝑙𝑢𝑒 𝑡 = −0.99 with 𝑑. 𝑓. = 9 is between 0.250 and 0.125
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.01, we do not reject 𝐻0
6. A study of fox rabies in southern Germany gave the following information about
different regions and the occurrence of rabies in each region. A random sample of
𝑛1 = 16 locations in region I gave the following information about the number of
cases of fox rabies near that location.
𝒙𝟏 Region I Data
1
8
8
8
7
8
8
1
3
3
3
2
5
1
4
6
A second random sample of 𝑛2 = 15 locations in region II gave the following
information about the number of cases of fox rabies near the location.
𝒙𝟐 Region II Data
1
1
3
1
4
8
5
4
4
2
2
5
6
9
4
Does this information indicate that there is a difference ( either way ) in the mean
number of cases of fox rabies between the two regions ? Use 𝛼 = 0.05
Using a calculator
𝑥̅1 = 4.75 , 𝑠1 = 2.82 , 𝑛1 = 16
𝑥̅2 = 3.93 , 𝑠2 = 2.43 , 𝑛2 = 15
𝐻0 ∶ 𝜇1 = 𝜇2
𝑡=
𝑥̅1 −𝑥̅ 2
𝑠2 𝑠2
√ 1+ 1
𝑛1 𝑛2
𝐻1 ∶ 𝜇1 ≠ 𝜇2
=
4.75−3.93
=
2
2
√2.82 +2.43
16
15
𝛼 = 0.05
0.82
√
=
0.497+0.394
0.82
√0.891
= 0.89
𝑃 − 𝑣𝑎𝑙𝑢𝑒 𝑡 = 0.89 with 𝑑. 𝑓. = 14 and two – tailed is between 0.500 and 0.250
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we do not reject 𝐻0
7. Would you prefer spending more federal tax money on the arts? Of a random
sample of 𝑛1 = 220 women, 𝑟1 = 59 responded yes. Another random sample of 𝑛2 =
175 men showed that 𝑟2 = 56 responded yes. Does this information indicate a
difference ( either way ) between the population proportion of women and the
population proportion of men who favor spending more tax dollars on the arts ? Use
𝛼 = 0.05
𝑟1 = 59 , 𝑛1 = 220 , 𝑟2 = 56 , 𝑛2 = 175
𝐻0 ∶ 𝑝1 − 𝑝2 = 0
𝐻1 ∶ 𝑝1 ≠ 𝑝2
𝛼 = 0.05
𝑝̅ =
𝑟1 +𝑟2
𝑛1 +𝑛2
220+175
𝑟1
59
𝑝̂1 =
𝑧=
59+56
=
𝑛1
=
220
𝑝̂1 − 𝑝̂2
𝑝̅ 𝑞̅ 𝑝̅ 𝑞̅
√
𝑛1 + 𝑛2
=
= 0.27
=
115
395
= 0.291
𝑝̂2 =
𝑟2
𝑛2
=
𝑞̅ = 1 − 𝑝̅ = 1 − 0.291 = 0.709
56
175
= 0.32
−0.05
√(0.291)(0.32) + (0.291)(0.32)
220
175
=
𝑝̂1 − 𝑝̂2 = 0.27 − 0.32 = −0.05
−0.05
√0.0004 + 0.0005
=
−0.05
√0.0009
= −1.67
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.67) = 2(0.0475) = 0.095
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 > 0.05, we fail to reject 𝐻0
8. A random sample of 𝑛1 = 228 voters registered in the state of California showed
that 141 voted in the last general election. A random sample of 𝑛2 = 216 registered
voters in the state of Colorado showed that 125 voted in the most recent general
election. Do these data indicate that the population proportion of voter turnout in
Colorado is higher than that in California? Use a 5% level of significance.
𝛼 = 0.05
𝑟1 = 114 , 𝑛1 = 228 , 𝑟2 = 125 , 𝑛2 = 216
𝐻0 ∶ 𝑝1 − 𝑝2 = 0
𝐻1 ∶ 𝑝1 < 𝑝2
𝛼 = 0.05
𝑝̅ =
𝑟1 +𝑟2
=
114+125
𝑛1 +𝑛2
228+216
𝑟1
114
𝑝̂1 =
𝑛1
=
228
𝑝̂1 − 𝑝̂2
𝑧=
𝑝̅ 𝑞̅ 𝑝̅ 𝑞̅
𝑛1 + 𝑛2
√
=
= 0.50
=
239
444
= 0.538
𝑝̂2 =
𝑟2
𝑛2
=
𝑞̅ = 1 − 𝑝̅ = 1 − 0.538 = 0.462
125
216
= 0.58
−0.08
√(0.538)(0.462) + (0.538)(0.462)
228
216
𝑃 − 𝑣𝑎𝑙𝑢𝑒 = 2𝑃(𝑧 < −1.67) = 0.0475
Since 𝑃 − 𝑣𝑎𝑙𝑢𝑒 ≤ 0.05, we reject 𝐻0
𝑝̂1 − 𝑝̂2 = 0.50 − 0.58 = −0.08
=
−0.08
√0.0011 + 0.0012
=
−0.08
= −1.67
0.048