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 7 = 205.8mg/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

Dear student, we can provide you with solution of one type of question at a time.

Calcium level vries with

u = 9.5, 0 = 0.4

n= 180

Sample size is large enough n>30 to assume normal distribution using Central Limit Theorem

a)  $CI = \overline{x}\pm Z_{\frac{\alpha }{2}}(\frac{\sigma}{\sqrt{n}})$

for 90% confidence ( using excel formula =-NORM.S.INV(0.05))

Ze = 1.645 2

$CI = 9.58\pm 1.645(\frac{0.4}{\sqrt{180}})$

$CI = 9.58\pm 0.049$

The lower limit of the confidnce interval calculated above is greater than 9.5, which is a clear evidence that the mean calcium level from which these women come is greater than 9.5.

b) Margin of error = $ME= 1.645(\frac{0.4}{\sqrt{180}})=0.049$

c) If you need to half the margin of error you roughly nedd 4 times the sample size now using.

By calculation

for ME = $\frac{0.049}{2}=0.0245$

for 90% confidence

Ze = 1.645 2

$n=(\frac{1.645*0.4}{0.0245})^2\approx 722 (rounded.up))$ ( the difference is because of rounding of numbers in between the calculation)

I HOPE IT HELPED :)