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 78 newborns whose mothers smoked during pregnancy has mean bone mineral content != 0.089g/cm and standard deviation ! = 0.022g/cm; a sample of 115 infants whose mothers did not smoke has mean != 0.083g/cm and standard deviation != 0.023g/cm. Assume that the underlying population variances are equal. a) Are the two samples of data 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

a) The samples are independent

b)
`H_(0)`: mu1=mu2*

`H_(a)`: mu1≠mu2

c) There is no significant evidence that an expectant mother’s cigarette smoking has any effect on the bone mineral content of her otherwise healthy child

Explanation:

Newborns whose mothers smoked during pregnancy and whose didn't smoke are two different populations. Therefore the samples are independent.

* Let mu1 be the mean bone mineral content of newborn babies whose mothers smoked during pregnancy

Let mu2 be the mean bone mineral content of newborn babies whose mothers didn't smoke during pregnancy

Null and alternative hypotheses are:

• 
  `H_(0)`: mu1=mu2

• 
  `H_(a)`: mu1≠mu2

Test statistic can be found using the equation:

`z=\frac{X-Y}{\sqrt{(s(x)^2)/(N(x))+(s(y)^2)/(N(y))}}` where

• X is the sample mean bone mineral content of newborn babies whose mothers smoked during pregnancy

• Y is the sample mean bone mineral content of newborn babies whose mothers didn't smoke during pregnancy

• s(x) is the sample standard deviation of bone mineral content of newborn babies whose mothers smoked during pregnancy

• s(y) is the sample standard deviation of bone mineral content of newborn babies whose mothers didn't smoke during pregnancy

• N(x) is the sample size for bone mineral content of newborn babies whose mothers smoked during pregnancy

• N(y) is the sample size for bone mineral content of newborn babies whose mothers didn't smoke during pregnancy

`z=\frac{0.089-0.083}{\sqrt{(0.022^2)/(78)+(0.023^2)/(115)}}` ≈1.825

two tailed p-value of test statistic is 0.068. Since 0.068>0.05, we fail to reject the null hypothesis. There is no significant evidence that an expectant mother’s cigarette smoking has any effect on the bone mineral content of her otherwise healthy child