Using Hypothesis testing,
a) Two samples are independent.
b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.
Hₐ : an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.
c) Null hypothesis is accepted.
so, an expectant mother's cigarette smoking has no effect on the bone mineral.
We have given that,
A study which is conducted for check an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.
For new-born whose mother's cigarette smoking
sample size , n₁= 77
X-bar, x₁-bar = 0.098 g/cm
standard deviations, s₁ = 0.026 g/cm
For new-born whose mother did not cigarette smoking
sample size , n₂ = 161
standard deviations, s₂ = 0.025 g/cm.
mean (X-bar) , x₂-bar = 0.095 g/cm
a) the two samples are independent since they are different types of mothers, smoking mothers and non smoking mothers
b)H₀: an expectant mother's cigarette smoking has any not effect on the bone mineral content of her otherwise healthy child.
Hₐ: an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child.
c) Test statistic:
t = (x₁-bar - x₂-bar )/S (√1/n₁+1/n₂)
where , S = √(s₁²(n₁ - 1) + s₂²(n₂ -1))/n₁+n₂ - 2
S = √((0.026)²(76) +(0.025 )²(160))/236
= 0.0253
then, t = (0.098 - 0.095 )/0.0253(√1/77 +1/161 )
=> t = 0.859
Using the critacal table critical t is
Critical t = ±1.970065
Degrees of freedom =236.0000
P-Value=0.3935 which is greater than α(0.05),
So , we accept H₀
Thus, an expectant mother's cigarette smoking has not effect on the bone mineral content of her otherwise healthy child.
To learn more about Hypothesis testing, refer:
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