1. Two machines are used to fill plastic bottles with dishwashing detergent. The
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Question
1. Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volumes are known to be 1= 0.10 and 2= 0.15 fluid ounces for the two machines, respectively. Two random samples of n1= 12 bottles from machine 1 and n2= 10 bottles from machine 2 are selected, and the sample mean fill volumes are 30.61 fluid ounces from machine 1 and 30.34 fluid ounces for machine 2. a) Test the hypothesis that both machines fill to the same mean volume. Use =0.05. b) What is the P-value? Interpret this value. c) Construct a 90% two-sided CI on the mean difference. Interpret this interval. d) What is the -error of the test if the true difference in mean fill volumes is 1.2 fluid ounces? 1. Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volumes are known to be 1= 0.10 and 2= 0.15 fluid ounces for the two machines, respectively. Two random samples of n1= 12 bottles from machine 1 and n2= 10 bottles from machine 2 are selected, and the sample mean fill volumes are 30.61 fluid ounces from machine 1 and 30.34 fluid ounces for machine 2. a) Test the hypothesis that both machines fill to the same mean volume. Use =0.05. b) What is the P-value? Interpret this value. c) Construct a 90% two-sided CI on the mean difference. Interpret this interval. d) What is the -error of the test if the true difference in mean fill volumes is 1.2 fluid ounces? 1. Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volumes are known to be 1= 0.10 and 2= 0.15 fluid ounces for the two machines, respectively. Two random samples of n1= 12 bottles from machine 1 and n2= 10 bottles from machine 2 are selected, and the sample mean fill volumes are 30.61 fluid ounces from machine 1 and 30.34 fluid ounces for machine 2. a) Test the hypothesis that both machines fill to the same mean volume. Use =0.05. b) What is the P-value? Interpret this value. c) Construct a 90% two-sided CI on the mean difference. Interpret this interval. d) What is the -error of the test if the true difference in mean fill volumes is 1.2 fluid ounces?Explanation / Answer
PART A & B.
Given that,
mean(x)=30.61
standard deviation , 1 =0.1
number(n1)=12
y(mean)=30.34
standard deviation, 2 =0.15
number(n2)=10
null, Ho: u1 = u2
alternate, H1: 1 != u2
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=30.61-30.34/sqrt((0.01/12)+(0.0225/10))
zo =4.86
| zo | =4.86
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =4.862 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 4.86 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: 4.86
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
PART C.
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ ( 30.61-30.34) ±Z a/2 * Sqrt( 0.01/12+0.0225/10)]
= [ (0.27) ± Z a/2 * Sqrt( 0.0031) ]
= [ (0.27) ± 1.645 * Sqrt( 0.0031) ]
= [0.1787 , 0.3613]
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