The data in table 1 below are measurements from a group of 10 normal males and 1
ID: 3204834 • Letter: T
Question
The data in table 1 below are measurements from a group of 10 normal males and 11 males with left-heart disease taken at autopsy at a particular hospital. Measurements were made on several variables at that time., and the table presents the measurements on total heart weight (THW) and total body weight (BW). Assume that the diagnosis of left-heart disease is made independently of these variables. Using the data in table 1 construct a 95% confidence interval for the mean of body weight for normal males. Using the data in table 1 construct a 90% confidence interval for the variance of total heart weight for left-heart disease males. Using the data in table 1 construct a 99% confidence interval for the variance of total heart weight for normal males. Are the variances of total heart weight (THW) in the two groups (Left-heart disease males and normal males) same at a significant alpha 0.05? Based on the result from 1A which t-test (equal or unequal variances) is used to test if the means of total heart weight (THW) the same or not at a significant alpha 0.05 by two-sided test? Implement the test and report a p-value (two-sided)?Explanation / Answer
1.1 For small sample size, n<30, and known sample standard deviation, use 1-sample t distribution.
The 95% c.i=xbar+-tdf=n-1, alpha/2(s/sqrt n), where, xbar is sample mean, s is sample standard deviation, and n is sample size, t denotes t critical at n-1 df and alpha/2. Compute xbar and s by using formulas, xbar=sigma x/n, s=sqrt[1/n-1 sigma (x-xbar)^2].
=56.10+-2.26(11.31/sqrt 10)
=(48.01, 64.19) [ans]
1.2 The 90% c.i for variance is as follows:
(n-1)s^2/X^2alpha/2<=sigma^2<=(n-1)s^2/X^21-alpha/2 , compute variance using formula, s^2=1/n-1 sigma (x-xbar)^2, where, xbar=sigma x/n, where, x denotesheart weight for left heart disease males and n is sample size, X^2 denote chi-square critical value at alpha/2, and 1-alpha/2 respectively.
10*19020/18.3070<=sigma^2<=10*19020/3.9403 [alpha=0.10, therefore, alpha/2=0.05, and 1-alpha/2=0.95]
10389.47<=sigma^2<=48270.44 (ans)
1.3
The 99% c.i for variance is as follows:
(n-1)s^2/X^2alpha/2<=sigma^2<=(n-1)s^2/X^21-alpha/2 , compute variance using formula, s^2=1/n-1 sigma (x-xbar)^2, where, xbar=sigma x/n, where, x denotes heart weight for normal males and n is sample size, X^2 denote chi-square critical value at alpha/2, and 1-alpha/2 respectively.
9*2084/23.5894<=sigma^2<=9*2084/1.7349 [alpha=0.01, therefore, alpha/2=0.005, and 1-alpha/2=0.995]
795.1029<=sigma^2<=10810.9978 (ans)
1.4 Hypotheses are as follows:
H0:sigma1^2=sigma2^2 (population variances of two groups are equal)
H1: sigma1^2=/=sigma2^2 (population variances of two groups are not equal)
alpha=0.05.
Test statistic:
F=s1^2/s2^2, where, s1 and s2 are sample standard deviations of group1 (left heart disease males, and normal males respectively). Compute, sample standard deviation by performing sqaure root of variance.
=19020/2084
=9.12
p value for F(9,10)=9.12 is 0.0009.
Conclusion: Per rule, reject H0, if p value is less than alpha=0.05. Here, p value is less than 0.05, therefore, reject H0 and conclude that there is significant difference in variance of total heart weight of two groups.
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