In: Computer Science

**Construct a 95% confidence interval for an experiment that found the sample mean temperature for a certain city in August was 101.82, with a population standard deviation of 1.2. There were 6 samples in this experiment.**

**A group of 10, foot - surgery patients had a mean weight of 240 pounds. The sample standard deviation was 25 pounds. Find a confidence interval for a sample for the true mean weight of all foot – surgery patients. Find a 95% Confidence Interval.**

**510 people applied to the Bachelor’s in Elementary Education program at Florida State College. Of those applicants, 57 were men. Find the 90% Confidence Interval of the true proportion of men who applied to the program.**

**A study revealed that 65% of men surveyed supported the war in Afghanistan and 33% of women supported the war. If 100 men and 75 women were surveyed, find the 90% confidence interval for the data’s true difference in proportions.**

1)

Confidence interval(CI) for normal distribution is (-z*/, +z*/)

here =101.82

n=6, =1.2

and Z value will be 1.96 which corresponds to (100-95)/2 %= 2.5%

z*/= 0.96

Therefore CI will be (101.82-0.96,101.82+0.96)= (100.86,102.78)

2)

Here we will use a t-distribution because n<30 and population mean is given as well .

Confidence interval(CI) for t-distribution is (-t*s/, +t*s/)

here =240

n=10, s=25

df=n-1 =10-1=9

By looking up in t-distribution table we find the value for df=9 and 2.5% value we get t=2.262

t*s/= 17.88

Therefore CI will be (240-17.88,240+17.88)= (222.12,257.88)

3)

CI for population proportion is( , )

n=510

Here alpha=1-0.9 =0.1

Z value for alpha/2 (.1/2=.05) is 0.13 ( By looking in the z table)

=57/510=0.112

1-=1-.112=0.888

=0.00182

Therefore CI will be (0.112-0.00182,0.112+0.00182)=(0.11018,0.11382)

4)

CI for difference in proportions is

( , )

n1=100,n2=75

=0.65

=0.33

= 1-0.65=0.35

=1-.33=0.67

Since it is the same 90% confidence interval as in previous sum z value will be 0.13

=0.0098

=0.65-0.33=0.32

Therefore CI will be (0.32-0.0098,0.32+0.0098)=(31.9902,32.0098)

Forma a 95% confidence interval on the mean pouring temperature
for iron castings if a sample of size 25 yielded a mean of 2560 and
a standard deviation (s) of 20.

Construct the 95% confidence interval for AOret using its the
mean
and standard deviation. Construct another 95% confidence interval
for AOret
using the mean of AOret and the average AOSD. Explain which one is
more
reliable.
Date
AllOrd
Aoret
AOSD
AUDUSD
AUDSD
TV (m)
IR (%)
MV
PE
2000/4/28
3085.1
0.0703
0.5909
0.0253
5122.6
5.72
634878
22.08
2000/5/31
3040.6
-1.44%
0.0451
0.5735
0.0358
6373.5
5.98
616440
21.36
2000/6/30
3257.6
7.14%
0.0337
0.5986
0.0383
7439.0
6
661306
22.74
2000/7/31
3213.6
-1.35%...

Problem:
Construct and interpret a 90%, 95%, and 99% confidence interval for
the mean heights of either adult females or the average height of
adult males living in America. Do not mix genders in your sample as
this will skew your results. Gather a random sample of size 30 of
heights from your friends, family, church members, strangers, etc.
by asking each individual in your sample his or her height. From
your raw data convert individual heights to inches....

Refer to the accompanying data set and construct a 95%
confidence interval estimate of the mean pulse rate of adult
females; then do the same for adult males. Compare the results.
Males : 86. 76. 49. 61. 55. 63. 53. 73. 53. 61. 74. 59. 63. 76.
85. 68. 63. 94. 41. 84. 74. 64. 69. 72. 55. 68. 58. 78. 69. 66. 64.
94. 56. 66. 60. 57. 68. 72. 87. 60.
Females: 81. 91. 60. 68. 57. 81....

Assuming that the population is normally distributed, construct
a 95% confidence interval for the population mean for each of the
samples below. Explain why these two samples produce different
confidence intervals even though they have the same mean and
range.
Sample A 1 1 2 4 5 7 8 8
Sample B 1 2 3 4 5 6 7 8

Assuming that the population is normally distributed, construct
a 95%
confidence interval for the population mean, based on the
following sample size of n equals 7. 1, 2, 3,
4, 5, 6, 7, and 23
In the given data, replace the value
23
with
7
and recalculate the confidence interval. Using these results,
describe the effect of an outlier (that is, an extreme value) on
the confidence interval, in general.
Find a 95%
confidence interval for the population mean, using...

Refer to the accompanying data set and construct a 95%
confidence interval estimate of the mean pulse rate of adult
females; then do the same for adult males. Compare the results.
Males
Females
84
80
77
97
52
60
59
65
51
55
59
79
52
78
75
85
50
86
65
59
70
34
62
68
65
85
76
74
79
75
65
63
68
68
99
81
44
62
86
64
70
82
67
85
70
68
71...

Refer to the accompanying data set and construct a 95%
confidence interval estimate of the mean pulse rate of adult
females; then do the same for adult males. Compare the results.
Males Females
84 78
71 95
49 56
63 64
53 54
61 82
51 81
75 88
54 89
62 57
69 36
59 65
62 86
78 74
80 75
65 64
65 68
94 77
45 61
86 61
71 82
63 83
74 68
74 ...

Assuming that the population is normally distributed, construct
a 95% confidence interval for the population mean, based on the
following sample size of n=6.
1, 2, 3, 4, 5,and 15
In the given data, replace the value 15 with 6 and recalculate
the confidence interval. Using these results, describe the effect
of an outlier (that is, an extreme value) on the confidence
interval, in general.
Find a 95% confidence interval for the population mean, using
the formula or technology.

Refer to the accompanying data set and construct a 95%
confidence interval estimate of the mean pulse rate of adult
females; then do the same for adult males. Compare the results.
Males Females
84 81
72 97
51 56
62 67
55 54
64 81
51 78
79 87
53 86
63 56
71 37
59 67
63 84
81 77
85 78
64 61
65 67
96 81
45 58
89 65
69 85
64 81
71 70
70 ...

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