Business Statistics (BUS 505) Assignment 9 Answer the question: 1 Given that, ; ; (a) 95% confidence interval is: 100(1 - ) = .95 = .05 So, we know that, = [ ] = 2.72< <3.08 So, 95% confidence interval range is from 2.7 to 3.08 (b) The probability content associated the interval from 2.81 to 2.99 is: [w = 2.99 – 2.81 = .18] = = 1 = .8413 So, 1 - = .8413 = .1587 = .3174 Now 1 - = 1 - .3174 = .6826 or 68.26% The probability content associated the interval from 2.81 to 2.99 is 68.26% Answer the question: 2 2. Given that, (a) 99% confidence interval is: 100(1 - ) = .999 = .01 So, we know that, Page 1 of 20
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The 95% Confidence interval for the population mean
[ ]
= 3.92- <μ< 3.92+
= 3.92- .0779 <μ< 3.92 + .079
= 3.84 <μ< 3.99
So, 95% confidence interval range is from 3.84 to 3.99
Answer the question: 6 Given that, n = 541 =3.81 Sx =1.34
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Business Statistics (BUS 505) Assignment 9
The 90% Confidence Interval for the population mean100 %( 1-α) =90% α = .10 /2=.05 Z /2=1.65
(a) So, we know that, [ ]
= 3.81- <μ<3.81+
=3.71<μ<3.90
The 90% Confidence Interval for the population mean 3.71 to 3.91
(b) It will be narrower.
Answer the question: 7
We know, Given that, W=.2 Sx =1.045
So, .2= 2Z α/2(1.045) √457 => Z α/2=2.04=.9793 = >1- α/2=.9793 = > α/2=1-.9793 α=.0414 Or, 1- α=.9586 or, 95.86%
So, The Confidence Interval is 95.86 %.
Answer the question: 8Here, n=352 Sx=11.28 =60.41
= [ ]
=59.42 <μ< 61.40
Comment: Here, we see that, if 57% to more mark than they are adequate understanding the material. So we can say that students are adequate understanding of the written material.
Answer the question: 9
Here, n=174 Sx=1.43 =6.06 W=.2
(a) We know,
So, .2=2Z α/2(1.43) √174 => Z α/2=.92=.8212 = >1- α/2=.8212
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Business Statistics (BUS 505) Assignment 9
= > α/2=1-.8212 α=.3576 Or, 1- α=.6424 or, 64.24%
So, The Confidence Interval is 64.24 %.
(b) Comment: Here confidence is decrease that different factor exiting. Such as sample size and sample standard deviation are difference compare than exercise 7.
Based on the material of section 8.4 .I have to follow the t distribution. But in t distribution I have to calculate the degree of freedom. The degree of freedom is n-1 but here n = 1 so degree of freedom is 0. Because of the degree of freedom is 0 it’s not possible to find confidence interval for the population mean.
So, the confidence interval for population standard deviation is
= 3.92 < < 8.84
Confidence interval for population standard deviation range is from 3.92 to 8.84 of weight looses for patients of the clinic’s weight reduction program.