3040 CEE Fall 2008
CEE 304 Section 9 Handout (Two Sample Hypothesis Tests)
1. Are men taller than women as is often believed? Look at the following data on heights (with units of inches) that was collected by former CEE 304 students.
N 4 12 Mean 65.146 71.258 Median 64.500 71.000 Stdev 2.441 2.889
Women Men
Let us consider these students to be a random sample of Cornell engineering students. Use a t-test appropriate for small n. Womens height x = 65.146 sx = 2.441 nx = 4
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Mens height y = 71.258 sy = 2.889 ny = 12 T= ( X Y ) (x y )
2 2 sx s y + nx n y __ __ __
x, y UNKNOWN, small samples Use t-test,
(i)
What are the appropriate hypotheses? Null Hypothesis Ho: x = y Alternative Hypothesis: Ha: x < y lower-tailed test
(ii) (iii) At the 1% level, what would be the rejection region for the test? =1%
2 2 sx s y + n ny Degrees of freedom: = x =6 4 2 s 4 n2 sx nx y y + nx 1 n y 1 2
t , = t 0.01, 6 = -3.143 So, Ho is rejected if t test < t , = -3.143 (lower tailed t test).
(iv) Approximately, what is the p-value for this data? 1
CEE 3040 Fall 2008 For lower tailed t test, P-value = P( T < t test ) T= ( X Y ) ( x y )
2 2 sx s y + nx n y __ __
t test
= - 4.135
From student-t table A.8, P-value = P( T < 4.135 | = 6 ) = 0.004 (v) What do you conclude at the 1% level? =1% (1) t test = 4.135 < t , = 3.143 So, Ho is rejected at level 1% (2) = 0 . 0 1 > P-value = 0.004 So, Ho is rejected at level 1% (vi) If the mean of women and men is really 65 and 71, respectively, and the true standard deviations are both 2.5, what is the type II error for a 1% test? Type II Error : ne-tail test, = 1 %, df = = 6 d= a 0
2 2 x y = 1.571 + nx n y
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Figure Chart A.17, = 0.2
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CEE 3040 Fall 2008 2. Groundwater Example: You concerned are with nitrate concentrations at wells upstream and downstream of a landfill. The following data was collected at the two locations. Downstream x = 5.6 mg/L sx = 2.1 mg/L nx = 11
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Upstream y = 4.2 mg/L sy = 1.8 mg/L ny = 6 T= ( X Y ) (x y )
2 2 sx s y + nx n y __ __ __
x, y UNKNOWN, small samples Use t-test,
(i)
What are the appropriate hypotheses? Null Hypothesis Ho: x = y Alternative Hypothesis: Ha: x > y upper-tailed test At the 5% level, what would be the rejection region for the test? =5%
2 2 sx s y + n ny Degrees of freedom: = x = 12.06 12 4 2 s 4 n2 sx nx +y y nx 1 n y 1 2
(ii)
t , = t 0.05,12 = 1.782 For upper tailed t test, Ho is rejected if t test > t , = 1.782. (iii) Approximately, what is the p-value for this data? For upper tailed t test, P-value = P( T > t test ) T= ( X Y ) ( x y )
2 2 sx s y + nx n y __ __
t test
= 1.44
So, P-value = P( T > 1.44 | = 12 ) = 0.087
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CEE 3040 Fall 2008 (iv) What do you conclude at the 5% level? =5% (1) t test = 1.44 < t , = 1.782 we fail to reject Ho at level 5% (2) = 0 . 0 5 < P-value = 0.087 Ho is NOT rejected at level 5% (v) If the mean downstream and upstream is really 5.5 and 4.0, respectively, and the true standard deviations are both 2.0, what is the power of the test (i.e. the probability we correctly reject the null hypothesis) given = 0.05? Type II Error : ne-tail test, = 5 %, df = = 12 d= a o
2 x2 y = 0.41 + nx n y
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Figure Chart A.17, = 0.62 Power of test 1- = 1 - 0.62 = 0.38 (vi) Construct a 90% confidence interval for the difference between the true means x and y of the downstream and upstream concentrations. = 0 . 1 , t / 2, = t 0.05,12 = 1.782 90% CI: X Y t / 2,
(
)
2 2 sx s y + = [-0.329, 3.129] nx n y
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