1. Two machines are used to fill plastic bottles with dishwashing detergent. The
ID: 3360175 • Letter: 1
Question
1. Two machines are used to fill plastic bottles with dishwashing detergent. The standard deviations of fill volumes are known to be ,-0.10 and 0.15 fluid ounces for the two machines, respectively. Two random samples of n 12 bottles from machine 1 and n- 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
a)
Given that,
mean(x)=30.61
standard deviation , 1 =0.1
number(n1)=12
y(mean)=30.64
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=30.61-30.64/sqrt((0.01/12)+(0.0225/10))
zo =-0.54
| zo | =0.54
critical value
the value of |z | at los 0.05% is 1.96
we got |zo | =0.54 & | z | =1.96
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.54 ) = 0.58901
hence value of p0.05 < 0.58901,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: 1 != u2
test statistic: -0.54
critical value: -1.96 , 1.96
decision: do not reject Ho
b)p-value: 0.58901
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 = [ ( 30.61-30.64) ±Z a/2 * Sqrt( 0.01/12+0.0225/10)]
= [ (-0.03) ± Z a/2 * Sqrt( 0.0031) ]
= [ (-0.03) ± 1.96 * Sqrt( 0.0031) ]
= [-0.1388 , 0.0788]
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interpretations:
1. we are 95% sure that the interval [-0.1388 , 0.0788] contains the difference between
true population mean U1 - U2
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the difference between
true population mean U1 - U2
3. Since this Cl does n't contain a zero we can conclude at 0.05 true mean
difference is not zero.
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