11) please help A study was done on skull sizes of humans during different time
ID: 3219770 • Letter: 1
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11) please help
A study was done on skull sizes of humans during different time periods. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. The test statistic, t, is (Round to two decimal places as needed.) The P-value is (Round to three decimal places as needed.) State the conclusion for the test. A. Reject the null hypothesis. There is not sufficient evidence to support the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150 B. Fail to reject the null hypothesis. There is sufficient evidence to support the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150. C. Reject the null hypothesis. There is sufficient evidence to support the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150. D. Fail to reject the null hypothesis. There is not sufficient evidence to supped the claim that the mean skull breadth was less in 4000 B.C. than A.D. 150. b. Construct a confidence interval suitable for testing the claim that the mean skull breadth in 4000 B.C is less than the mean skull breadth in A.D 150.Explanation / Answer
(a) Data:
n1 = 27
n2 = 27
x1-bar = 131.27
x2-bar = 132.55
s1 = 5.15
s2 = 5.08
Hypotheses:
Ho: 1 ³ 2
Ha: 1 < 2
Decision Rule:
= 0.05
Degrees of freedom = MIN((27 - 1), (27 - 1)) = 26
Critical t- score = -1.705617901
Reject Ho if t < -1.705617901
Test Statistic:
SE = {(s1^2 /n1) + (s2^2 /n2)} = (((5.15^2)/27) + ((5.08^2)/27)) = 1.392159261
t = (x1-bar -x2-bar)/SE = (131.27 - 132.55)/1.39215926079145 = -0.92
(b) p- value = 0.183
(c) Option D
(d)
n1 = 27
n2 = 27
x1-bar = 131.27
x2-bar = 132.55
s1 = 5.15
s2 = 5.08
% = 95
Degrees of freedom = n1 + n2 - 2 = 27 + 27 -2 = 52
Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((27 - 1) * 5.15^2 + ( 27 - 1) * 5.08^2)/(27 + 27 -2)) = 5.115119744
SE = Pooled s * ((1/n1) + (1/n2)) = 5.11511974444392 * ((1/27) + (1/27)) = 1.392159261
t- score = 2.006646761
Width of the confidence interval = t * SE = 2.00664676070409 * 1.39215926079145 = 2.793571871
Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = -1.28 - 2.79357187105137 = -4.073571871
Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = -1.28 + 2.79357187105137 = 1.513571871
The 95% confidence interval is [-4.074, 1.514]
(e) Yes (since the confidence interval contains 0)
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