C. Find the length of the confidence interval
Spencer Marvin P. Esguerra
Answered question
2022-06-20
C. Find the length of the confidence interval given the following data:(s – sample
standard deviation)
1. 𝑠 = 2.36, n = 350, confidence level: 99%
2. 𝑠 = 3.35, n = 250, confidence level: 99%
3. 𝑠 = 3, n = 275, confidence level: 95%
4. 𝑠 = 6, n = 425, confidence level: 98%
5. 𝑠 = 9, n = 501, confidence level: 99%
Answer & Explanation
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Skilled2022-09-14Added 403 answers
1. The given data are as follows,
Sample standard deviation (s) is 2.36
Sample size (n) is 350
Confidence level is 99%
Level of significance level ()
= 100-99
= 1%
= 0.01
p-value
= [1-(/2)]
= [1-(0.01/2)]
= 0.995
From standard normal table,
Critical value z at 0.995 is 2.5762.58
Error margin
= {*[s/sqr root(n)]}*2
= {2.58*[2.36/sqr root(350)]}*2
= 0.32546005*2
= 0.6509
Therefore confidence interval is 0.6509
2. The given data are as follows,
Sample standard deviation (s) is 3.35
Sample size (n) is 250
Confidence level is 99%
Level of significance level ()
= 100-99
= 1%
= 0.01
p-value
= [1-(/2)]
= [1-(0.01/2)]
= 0.995
From standard normal table,
Critical value z at 0.995 is 2.5762.58
Error margin
= {*[s/sqr root(n)]}*2
= {2.58*[3.35/sqr root(250)]}*2
= 0.546631316*2
= 1.0933
Therefore confidence interval is 1.0933
3. The given data are as follows,
Sample standard deviation (s) is 3
Sample size (n) is 275
Confidence level is 95%
Level of significance level ()
= 100-95
= 5%
= 0.05
p-value
= [1-(/2)]
= [1-(0.05/2)]
= 0.975
From standard normal table,
Critical value z at 0.975 is 1.96
Error margin
= {*[s/sqr root(n)]}*2
= {1.96*[3/sqr root(275)]}*2
= 0.354577341*2
= 0.7092
Therefore confidence interval is 0.7092
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