1. The following data represent petal lengths (in cm) for independent random samples of two species of Iris.
Petal length (in cm) of Iris virginica: x_{1}; n_{1} = 35
5.1 | 5.9 | 6.1 | 6.1 | 5.1 | 5.5 | 5.3 | 5.5 | 6.9 | 5.0 | 4.9 | 6.0 | 4.8 | 6.1 | 5.6 | 5.1 |
5.6 | 4.8 | 5.4 | 5.1 | 5.1 | 5.9 | 5.2 | 5.7 | 5.4 | 4.5 | 6.4 | 5.3 | 5.5 | 6.7 | 5.7 | 4.9 |
4.8 | 5.9 | 5.1 |
Petal length (in cm) of Iris setosa: x_{2}; n_{2} = 38
1.5 | 1.9 | 1.4 | 1.5 | 1.5 | 1.6 | 1.4 | 1.1 | 1.2 | 1.4 | 1.7 | 1.0 | 1.7 | 1.9 | 1.6 | 1.4 |
1.5 | 1.4 | 1.2 | 1.3 | 1.5 | 1.3 | 1.6 | 1.9 | 1.4 | 1.6 | 1.5 | 1.4 | 1.6 | 1.2 | 1.9 | 1.5 |
1.6 | 1.4 | 1.3 | 1.7 | 1.5 | 1.7 |
(a) Use a calculator with mean and standard deviation keys to calculate x_{1}, s_{1}, x_{2}, and s_{2}. (Round your answers to two decimal places.)
x_{1} = | |
s_{1} = | |
x_{2} = | |
s_{2} = |
(b) Let μ_{1} be the population mean for
x_{1} and let μ_{2} be the
population mean for x_{2}. Find a 99% confidence
interval for μ_{1} − μ_{2}.
(Round your answers to two decimal places.)
lower limit | |
upper limit |
2. A study of parental empathy for sensitivity cues and baby temperament (higher scores mean more empathy) was performed. Let x_{1} be a random variable that represents the score of a mother on an empathy test (as regards her baby). Let x_{2}be the empathy score of a father. A random sample of 37 mothers gave a sample mean of x_{1} = 67.00. Another random sample of 27 fathers gave x_{2} = 61.04. Assume that σ_{1} = 10.92 and σ_{2} = 11.62.
(a) Let μ_{1} be the population mean of x_{1} and let μ_{2} be the population mean of x_{2}. Find a 95% confidence interval for μ_{1} – μ_{2}. (Use 2 decimal places.)
lower limit | |
upper limit |
Question 1
Part a)
X1 = 5.49
S1 = 0.56
X2 = 1.49
S2 = 0.22
Confidence interval :-
t(α/2, DF) = t(0.01 /2, 43 ) = 2.695
DF = 43
99% confidence interval is ( 3.72 , 4.26 )
Question 2
Confidence interval :-
Z(α/2) = Z (0.05 /2) = 1.96
Lower Limit = 0.34
Upper Limit = 11.58
95% Confidence interval is ( 0.34 , 11.58
)
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