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5.2
Here the difference in sample means = 11.5
Test statistic z0 = -1.88
p-value = 0.0601
a. Therefore, standard error of the difference in sample means = 11.5/1.88 = 6.117
b. This is a one sided test
c. If alpha = 0.05, we accept the null hypothesis of equality of population means since p-value (=
0.0601) is greater than 0.05.
d. A 955 confidence interval of the difference in sample means is given by: (sample difference in
means – Z0.025*standard error, sample difference in means + Z0.025*standard error).
Now, Z0.025 = 1.96. Therefore, the 95% confidence interval is given by: (-0.489, 23.48932)
5.8
Here we assume normality.
a. The 95% confidence interval of difference in means is given by: ((x1-bar – x2-bar) –
Z0.025*√(s1
2/n1 + s2
2/n2), (x1-bar – x2-bar) + Z0.025*√(σ1
2/n1 + σ2
2/n2)).
Therefore, 95% confidence interval of difference in mean is given by: (-4.47, -2.91)
b. Let, m1 = population mean of formulation 1
m2 = population mean of formulation 2
Here we want to test, H: m1 = m2 vs. K: m2 > m1
Here the test statistic is given by, Z = (x2-bar – x1-bar)/√ σ1
2/n1 + σ2
2/n2) which follows N(0,1)
under H. So we reject H at α = 0.05 if p-value of the test is less than 0.05. Now, p-value of the
test is given by, p-value = P(Z > observed Z) where Z follows N(0,1). Here observed Z = 9.225.
Therefore, -value = P(Z > 9.225) ≈ 0. Therefore, p-value of the test < 0.05 so we reject H at 55
level of confidence and conclude that the mean for formulation 2 is greater than mean of
formulation 1.
5.12
a. Sample mean for catalyst 1, x1-bar = 63.6 and sample mean for catalyst 2, x2-bar = 67.8
We know the standard deviation of active concentration is = 3
So, the 95% CI for difference in mean active concentration for the two catalysts = ((x1-bar – x2-bar) –
Z0.025*σ√(1/n1 + 1/n2), (x1-bar – x2-bar) + Z0.025*σ√(1/n1 + 1/n2)). Therefore the 95% CI is given by: (-6.83,
-1.57).
b. From the above results we can see that zero is not included in the 95% CI for mean difference.
Therefore, at 5% level of confidence we can conclude that there is significant evidence that the mean
active concentration depends on the choice of catalyst.
5.17
a. Here we assume that both populations are normally distributed and the same variance.
Suppose, m1 = population mean of single spindle saw
m2 = population mean of double spindle saw
Here we want to test, H: m1 = m2 vs. K: m1 ≠ m2
Here the test statistic is given by, t = (x1-bar – x2-bar)/s√(1/n1 + 1/n2) where x1-bar = sample mean for
single spindle saw, x2-bar = sample mean for double spindle saw, n1 = sample size for single spindle saw,
n2 = sample size for double spindle saw, s =((n1 – 1) s1
2 + (n2 – 1) s2
2)/(n1 + n2 – 2). Now, t follows t-
distribution with d.f. = n1 + n2 – 2. We reject H at 5% level of confidence if p-value of the test is less than
0.05.
Here, x1-bar = 66.385, x2-bar = 45.278, s1
2 = 7.895, s2
2 = 8.612, n1 = n2 = 15.
Therefore, observed t = 2.683.
Here the p-value of the test = 2P(t > 2.683) = 0.0122
Clearly, p-value of the test is less than 0.05 therefore we reject H at 5% level of significance.
b.The 95% CI of difference in mean is given by: ((x1-bar – x2-bar) – t* s√(1/n1 + 1/n2), (x1-bar – x2-bar) +
t* s√(1/n1 + 1/n2)) where t = 0.025 upper cut off point of t-distribution with d.f. = 28 (=2.048).
Therefore, the 95% CI is given by: (18.96, 23.26). Since the CI does not include zero we can conclude that
there is a significant difference in the means at 5% level of significance. Therefore, the conclusion using
CI is same as our previous conclusion.
5.19
a. Here we assume that both populations are normally distributed and the same variance.
Suppose, m1 = population mean of machine 1
m2 = population mean of machine 2
Here we want to test, H: m1 = m2 vs. K: m1 ≠ m2
Here the test...