Stats 2
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ADM 2304 E – Statistical Methods Assignment #1
Problem 1
Test and CI for Two Proportions: bmi_A, bmi_B
Null Hypothesis: No significant difference
Alternative Hypothesis: Significant difference
Event = 1
Variable
Sample p
bmi_A
192
1342
0.143070
bmi_B
97
720
0.134722
Difference = p (bmi_A) – p (bmi_B)
Estimate for difference = 0.143070 Ð- 0.134722 => 0.00834782
90% CI for difference = (-0.0178288, 0.0345244)
Test for difference = 0 (vs not = 0)
Z = 0.52 P-Value = 0.600
Therefore, based on these results we would not reject the null hypothesis. There is no significant difference in the weights of both cities.
Manual
Pooled P-hat
= (192 + 97) / (1342 + 720)
= 289 / 2062
= 0.140155
Z-stat
= 0.008348 / 0.016037
= 0.520546
Confidence Interval
= 0.008348 +/- 0.026179
Therefore this equals the interval of
(-0.0178288, 0.0345244)
The confidence interval calculated above is consistent with the conclusion in question A.
Problem 2
Test and CI for One Proportion: C3
Test of p = 0.5 vs p not = 0.5
Event = 1
Variable
X
Sample p 95% CI
Exact P-Value
1247
1937
0.643779
(0.621990, 0.665132)
0.000
b & c)
Test and CI for One Proportion: C4, C5, C6, C7, C8, C9, C10, C11,
Test of p = 0.5 vs p not = 0.5
Event = 1
Variable
Sample p
90% CI
(0.184473, 0.511179)
(0.537839, 0.853466)
(0.395195, 0.735240)
(0.588527, 0.889734)
(0.264760, 0.604805)
(0.395195, 0.735240)
(0.441309, 0.776082)
(0.350413, 0.693065)
(0.441309, 0.776082)
(0.395195, 0.735240)
(0.488821, 0.815527)
(0.537839, 0.853466)
(0.350413, 0.693065)
(0.537839, 0.853466)
(0.488821, 0.815527)
(0.641141, 0.924076)
(0.641141, 0.924076)
(0.441309, 0.776082)
(0.488821,