10.6 Two independent random samples are chosen from normal populations. The firs
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Question
10.6 Two independent random samples are chosen from normal populations. The first population has 1 = 5 and 1 = 3 and the second population has 2 = 4 and 2 = 6. For the first sample, n1 = 36 and the sample mean and standard deviation are 1 and S1 and for the second sample, n2 = 25 and the sample mean and standard deviation are 2 and S2. What is the distribution of each of the following? For each part give the type (name) of distribution and the parameters needed to specify the distribution. For example, U = the number of members in the first sample that are less than 5 is binomialwithn=36andp=.5.
(a) Xbar_1 (Hint: The variable has a normal distribution. Give its mean and variance.)
(b) (Xbar_1 - 5)/(S_1/6)
c) (Xbar_1 - 5)/(3/6)
d) 35*((S_1)^2) / 9
e) ((S_1)^2)/ 9/(((S_2)^2)/36)
f) Xbar_1 - Xbar_2
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.3
DF = 59
t = [ (x1 - x2) - d ] / SE
t = 0.769
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than - 0.77; that is, less than - 0.77 or greater than 0.77.
Thus, the P-value = 0.4444
Interpret results. Since the P-value (0.4444) is more than the significance level (0.05), we accept the null hypothesis.
f) The distribution of Xbar_1 - Xbar_2 is normal distribution.
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