Answer:
3.0 mi/h < σ < 8.54 mi/h
Step-by-step explanation:
Given:
Sample data: x: 65 62 62 55 62 55 60 59 60 70 61 68
Confidence = c = 98% = 0.98
To find:
Construct a 98% confidence interval estimate of the population standard deviation.
Solution:
Compute Mean:
number of terms in data set = n = 12
Mean = Sum of all terms / number of terms
= 65 + 62 + 62 + 55 + 62 + 55 + 60 + 59 + 60 + 70 + 61 + 68 / 12
= 739/12
Mean = 61.58
Compute standard deviation:
s = √∑(each term of data set - mean)/ sample size - 1
s = √∑([tex]_{x-} {\frac{}{x} }[/tex])²/n-1
= √( 65 - 61.58)² + (62 - 61.58)² + (62 - 61.58)² + (55 - 61.58)² + (62 - 61.58)² + (55 - 61.58)² + (60 - 61.58)² + (59 - 61.58)² + (60 - 61.58)² + (70 - 61.58)² + (61 -61.58)² + (68 - 61.58)² / 12-1
= √(11.6964 + 0.1764 + 0.1764 + 43.2964 + 0.1764 + 43.2964 + 2.4964 + 6.6564 + 2.4964 + 70.8964 + 0.3364 + 41.2164) / 11
= √222.9168/11
= √20.2652
= 4.50168
= 4.5017
s = 4.5017
Compute critical value using chi-square table:
For row:
degree of freedom = n-1 = 12 - 1 = 11
For Column:
(1 - c) / 2 = (1 - 0.98) / 2 = 0.02/2 = 0.01
1 - (1 - c) / 2 = 1 - (1-0.98) / 2 = 1 - 0.02 / 2 = 1 - 0.01 = 0.99
[tex]X^{2} _{1-\alpha/2}[/tex] = 3.053
[tex]X^{2} _{\alpha/2}[/tex] = 24.725
Compute 98% confidence interval of standard deviation:
[tex]\sqrt{\frac{n-1}{X^{2} _{\alpha/2}} } s[/tex] = [tex]\sqrt{\frac{12-1}{24.725} } ( 4.5017)[/tex] = [tex]\sqrt{\frac{11}{24.725} } ( 4.5017)[/tex] = [tex]\sqrt{0.44489}(4.5017)[/tex]
= 0.6670 (4.5017) = 3.0026
[tex]\sqrt{\frac{n-1}{X^{2} _{\alpha/2}} } s[/tex] = 3.0026
[tex]\sqrt{\frac{n-1}{X^{2} _{1-\alpha/2}} } s[/tex] = [tex]\sqrt{\frac{12-1}{3.053} } ( 4.5017)[/tex] = [tex]\sqrt{\frac{11}{3.053} } ( 4.5017)[/tex] = [tex]\sqrt{3.6030} (4.5017)[/tex]
= 1.8982 ( 4.5017) = 8.5449
[tex]\sqrt{\frac{n-1}{X^{2} _{1-\alpha/2}} } s[/tex] = 8.5449
3.0026 mi/h < σ < 8.5449 mi/h
