c. The condition for using normal distribution is:
np > 5 and n(1 – p) > 5
Therefore, the condition is satisfied.

200 * 0.25 = 50
200
0.75 = 150
4. a. Construct a 95% confidence interval:
23.491 to 27.509
25.5 ± 1.96(10.25 /
√100) = 25.5
±
2.009 = 23.491 to 27.509
b. Margin of error for 90% confidence interval:
1.686
= 1.645 (10.25 /
√100) = 1.686
c. Construct a 95% confidence interval for a random sample of 150:
23.86 to 27.14
25.5 ± 1.96(10.25 /
√150) = 25.5 ± 1.640 = 23.86 to 27.14
d. The length of the confidence interval decreases as the sample size increases.
5.
a. Construct a 99% confidence interval:
19.307 to 27.6927
23.5 ± 2.662 (12.2 /
√60) = 23.5 ± 4.1927
b. Margin for error on a 95% confidence interval:
3.15
= 2.001(12.2 /
√60) = 3.15
6.
a. With a 95% confidence interval what is the margin for error?
0.5118
2.064(1.24 / √25) =
0.5118
b. What is the 95% confidence interval estimate of the population mean score?
3.238
to
4.261
3.75
± 2.064 (1.24 /
√25) = 3.75 ± 0.5118 = 95% confidence interval.
7. a. Approximate mean = range / 4 = (55,000 – 35,000) / 4 =
5000
The planning value for the population standard deviation is
5000.
b. n = (z
α
/2)
2
(
σ)
2
/ E
2
= (1.96)
2
(5000)
2
/ 1000
2
=
96.04
The sample size should contain 96.04 or
97
university graduates if the margin for error is
100
c. n = (1.96)
2
(5000)
2
/ (500)
2
=
384.16

The sample size should contain 384.16 or
385
university graduates if the margin for effort
is 500
8.a. Construct a 95% confidence interval:
(0.4912, 0.5888)
p ± 1.96
√p(1-p)
/ √n = 0.54 ±
1.96 √0.54(0.46) / √400 = 0.048842948
0.54 + 0.048842948 = 0.5888
0.54 – 0.048842948 = 0.4912
(0.4912, 0.5888)
b. What is the margin of error for the 90% confidence interval?
0.0409932
Margin of Error
=
zα
/
2
√
p(1-p) /
√n
= 1.645
√(0.54)(0.46) / √400 =
0.0409932
c. Required sample size if desired margin of error is 3%, at the 95% confidence interval:
1061
(1.96
2
* 0.54(1 – 0.54) / 0.03
2
= 3.8416 * (0.54)(0.46) / 0.0009 =
1060.2816
9. a. Construct a 95% confidence interval to estimate the population proportion:
(0.1253,
0.1747)
p = x / n = 120 / 800 = 0.15
p ± 1.96
√p(1-p) / √n
0.15
± 1.96
√(0.15*0.85) / √800 = 0.024743787
0.15 + 0.024743787 = 0.1747
0.15 - 0.024743787 = 0.1253
(0.1253, 0.1747)
b. What is the margin of error of the estimate at the 99% confidence interval?
0.03252
E = 2.576*
√(0.15*0.85) / √800 =
0.03252406
10. How large should the sample size be?

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- Winter '15
- Business, Statistics, Normal Distribution, Standard Deviation