Question

5. The level of calcium in the blood in healthy young adults varies with a mean of about 9.5 mg per liter and standard deviation of about 0.4. A clinic in rural Guatemala measures the blood calcium of 180 healthy pregnant women at their first visit for prenatal care. The mean is 9.58. Is this an indication that the mean calcium level from which these women come is greater than 9.5? (Assume s = 0.4) a. Give a 90% confidence interval for the mean blood calcium level for this population of women and use it to answer the question. b. What is the margin of error above? c. Approximately how big would our sample need to be for the margin of error to be half as much with 90% confidence? 6. The blood glucose level a person varies throughout the day. Someone with a mean glucose level of more than 200 mg/dL is a diabetic. Diabetes is a serious medical condition and may need to be hospitalized or more extensive care. Sally has not been feeling well and her doctor suspects that she is a diabetic and so sends her home with a blood glucose monitor so that she can test her glucose levels over the course of a week. She was instructed to measure her glucose level 40 times that week. In so doing she found that her average blood glucose level was... a. Set up hypotheses for a significance test to determine if Sally is a diabetic. b. What would a Type I error mean in the context of this problem? c. What would a Type II error mean in the context of this problem? d. If T = 205.8 mg/dL and the standard deviation of her measurements was s = 10.6 mg/dL, what is the value of the test statistic that would be used for a significance test?

Answer 1

5.

TRADITIONAL METHOD
given that,
sample mean, x =9.58
standard deviation, s =0.4
sample size, n =180
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.4/ sqrt ( 180) )
= 0.03
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 179 d.f is 1.653
margin of error = 1.653 * 0.03
= 0.049
III.
CI = x ± margin of error
confidence interval = [ 9.58 ± 0.049 ]
= [ 9.531 , 9.629 ]
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DIRECT METHOD
given that,
sample mean, x =9.58
standard deviation, s =0.4
sample size, n =180
level of significance, α = 0.1
from standard normal table, two tailed value of |t α/2| with n-1 = 179 d.f is 1.653
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 9.58 ± t a/2 ( 0.4/ Sqrt ( 180) ]
= [ 9.58-(1.653 * 0.03) , 9.58+(1.653 * 0.03) ]
= [ 9.531 , 9.629 ]
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interpretations:
1) we are 90% sure that the interval [ 9.531 , 9.629 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
a.
90% sure that the interval [ 9.531 , 9.629 ] for the mean blood calcium level for this population of women
b.
margin of error = 0.049
c.
given data,
standard deviation =0.4
margin of error =0.5
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.1% LOS is = 1.645 ( From Standard Normal Table )
Standard Deviation ( S.D) = 0.4
ME =0.5
n = ( 1.645*0.4/0.5) ^2
= (0.658/0.5 ) ^2
= 1.732 ~ 2          
6.
Given that,
population mean(u)=200
sample mean, x =205.8
standard deviation, s =10.6
number (n)=40
null, Ho: μ=200
alternate, H1: μ>200
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.685
since our test is right-tailed
reject Ho, if to > 1.685
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =205.8-200/(10.6/sqrt(40))
to =3.461
| to | =3.461
critical value
the value of |t α| with n-1 = 39 d.f is 1.685
we got |to| =3.461 & | t α | =1.685
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 3.4606 ) = 0.00066
hence value of p0.05 > 0.00066,here we reject Ho
ANSWERS
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a.
null, Ho: μ=200
alternate, H1: μ>200
b.
Type 1 error is possible when it reject the null hypothesis, in this context
c.
Type 2 error is not possible because when its fails to reject the null hypothesis.
d.
test statistic: 3.461
critical value: 1.685
decision: reject Ho
p-value: 0.00066
we have enough evidence to support the claim that mean glucose level of more than 200mg/dL is a diabetic.