Recent studies have indicated that nicotine may suppress the growth of bone cell
ID: 3182110 • Letter: R
Question
Recent studies have indicated that nicotine may suppress the growth of bone cells and may be associated with decreased calcium absorption in humans. Since bones act as calcium stores, long periods of insufficient calcium can lead to bone loss and the risk of developing osteoporosis. Postmenopausal women are a group that is particularly at risk because of their loss of estrogen following menopause. A study was conducted to compare bone mineral density of post-menopausal women who had smoked throughout their adult lives with non-smoking post- menopausal women. It was hypothesized that the smoking group would have lower bone density than the non-smoking group. Low bone density indicates porous and fragile bones and potentially osteoporosis. Assume that bone density is approximately normally distributed. Summary statistics from the study are given in the table below.
a) Identify the random variable(s), of interest and their probability distribu- tion(s).
b) Test an appropriate set of hypotheses to determine if the bone density for the smoking group is more variable than that of the non-smoking group. Use = 0.05.
c) Test an appropriate set of hypotheses to determine if the study supports the researchers hypothesis. Use = 0.01.
d) Do the results of the study show that smoking can cause osteoporosis? Briefly explain your answer.
e) In the context of the problem,
i) State what it would mean to make a type I error.
ii) State what it would mean to make a type II error.
n Mean St.Dev Non-Smoking 23 1.68 .14 Smoking 19 1.47 .21Explanation / Answer
Given that,
mean(x)=1.68
standard deviation , 1 =0.14
number(n1)=23
y(mean)=1.47
standard deviation, 2 =0.21
number(n2)=19
null, Ho: u1 = u2
alternate, the bone density for the smoking group is more variable than that of the non-smoking groupH1: 1 > u2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=1.68-1.47/sqrt((0.0196/23)+(0.0441/19))
zo =3.73
| zo | =3.73
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =3.728 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail -Ha : ( p > 3.73 ) = 0.0001
hence value of p0.05 > 0.0001,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 > u2
test statistic: 3.73
critical value: 1.645
decision: reject Ho
p-value: 0.0001
We have evidence that the bone density for the smoking group is more variable than that of the non-smoking group
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