Question

A study was conducted to determine whether an expectant mother's cigarette smoking has any effect on the bone mineral content of her otherwise healthy child. A sample of 77 newborns whose mothers smoked during pregnancy has mean bone mineral content x-bar1 = 0.098 g/cm and standard deviation s1 = 0.026 g/cm; a sample of 161 infants whose mothers did not smoke has mean x-bar2 = 0.095 g/cm and standard deviation s2 = 0.025 g/cm. Assume that the underlying population variances are equal.
a. Are the two samples paired or independent?
b. State the null and alternative hypotheses of the two-sided test.
c. Conduct the test at the 0.05 level of significance. What do you conclude?

Answer 1

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:

brainly.com/question/4232174

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