Do a two-sample test for equality of means assuming unequal variances. Calculate
ID: 3297250 • Letter: D
Question
Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel. Comparison of GPA for randomly chosen college juniors and seniors: x_1 = 3.05, S_1 = 20, n_1 = 15, x_2 = 3.25, s_2 = 30, n_2 = 15, alpha = 025, left-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis Comparison of average commute miles for randomly chosen students at two community colleges: x_1 = 15, s_1 = 5, n_1 = 22, x_2 = 18, s_2 = 7, n_2 = 19, alpha = 05, two-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.) Based on the above data choose the correct decision. Reject the null hypothesis Do not reject the null hypothesis Comparison of credits at time of graduation for randomly chosen accounting and economics students: x_1 = 139, s_1 = 2.8, n_1 = 12, x_2 = 137, s_2 = 2.7, n_2 = 17, alpha = 05, right-tailed test. (Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places. Do not use "quick" rules for degrees of freedom.)Explanation / Answer
a-1 Degrees of freedom, df=(s1^2/n1+s2^2/n2)^2/{1/n1-1 (s1^2/n1)^2+1/n2-1 (s2^2/n2)^2}
Substitute values.
df=(0.2^2/15+0.2^2/15)^2/{1/15-1 (0.2^2/15)^2+1/15-1 (0.2^2/15)^2}
=28
t calculated=(x1bar-x2bar)/sqrt[s1^2/n1+s2^2/n2]
=(3.05-3.25)/sqrt[0.2^2/15+0.2^2/15]
=-2.7386
Using technology, the p value=0.0053
Look into t table for row (df=28) and column one tail (0.025). The t critical=-2.0484 (left tailed test).
a-2 Per rejection rule based on critical value, reject null hypothesis, if observed test statistic falls in critical region (observed test statistic is greater than or eqaul to critical value). Here, observed t falls in critical region. Therefore, reject null hypothesis. Per rejection rule, based on p value, reject null hypothesis if p value is less than alpha=0.05. Here, p value is less than 0.05, therefore, reject null hypothesis.
b-1 Substituting the values in the formula stated earlier to calculate degrees of freedom, the df=32
t-calculated=(15-18)/sqrt[5^2/22+7^2/19]
=-1.5564
p value=0.1295
t critical=+-2.0369 [look into t table with row=32 and column two tail 0.05]
b-2 Since, the test statistic does not fall in critical region (or the p value is not less than alpha=0.05), fail to reject null hypothesis. Ans>Do not reject null hypothesis.
c-1 Substituting given values in df formula, df=23
t =(139-137)/sqrt[2.8^2/12+2.7^21/7]
=1.9226
p value=0.0335
t-critical=1.7139 (right-tailed) [look iinto t table with row=23 and column 0.05 one tail.
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