Question

In: Statistics and Probability

A hospital wishes to justify the benefits of nutrition programs for pregnant women using birth weight...

A hospital wishes to justify the benefits of nutrition programs for pregnant women using birth weight data from newborns. The hospital hopes to show that the mean birth weight for newborns from mothers who complete the program is higher than the birth weight for newborns from mothers who do not complete the program. A group of 12 pregnant women were randomly divided into two groups; the first group received the nutrition program and the second group did not receive the program. The resulting weights (in grams) of the newborn babies from each group are shown below. Assume normality.
Group 1 Group 2
2574 2497
2572 2473
2512 2425
2672 2472
2605 2531
2649 2549
a.) What is the test statistic? Give your answer to four decimal places.
b.) What is the P-value associated with the test statistic? Give your answer to four decimal places.

Solutions

Expert Solution

Given
X1 bar 2597.333333 (AVERAGE()) X2 bar 2491.167
S1 57.91603117(STADEV()) S2 44.81257
n1 6 n2 6

Let alpha = 0.05

Equal variance test:

F = S1^2/S2^2 = 1.6703

F critical

FL​=0.14

FU​=7.146

FL​=0.14 < F=1.67 < FU​=7.146 ,DO not reject equal variance of null hypothesis

T test:

Hypothesis : α= 0.05
df 10 n1+n2-2
Ho: μ1​ = μ2
Ha: μ1​ > μ2
t Critical Value :
tc 1.812461123 T.INV(1-α,df) RIGHT
ts >= tc RIGHT To reject
Test :
Sp^2 2681.216667 ((n1-1)S1^2+(n2-1)S2^2)/(n1+n2-2)
t stat 3.551263144 (X1 bar-X2 bar )/SQRT(Sp*(1/n1 + 1/n2)) Equal vriance
P value :
P value 0.002628586 T.DIST.RT(ts,df) RIGHT
Decision :
P value < α Reject H0

P value < 0.05, Reject H0

There is enough evidence to conclude that the mean birth weight for newborns from mothers who complete the program is higher than the birth weight for newborns from mothers who do not complete the program.


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