Stat 200 Practice Test 3 Name— --- Sec ----
1. To test the claim that a vegetarian diet helped reduce cholesterol, the mean
cholesterol level of 46 seniors were recorded after six months of consuming this
diet. The researchers hypothesized that the mean cholesterol level after six months
should be less than the normal cholesterol value of 150. The sample average
cholesterol levels after the six month period was calculated to be 140 with a
standard deviation of 10.
a. With alpha=.01, answer the following, do a test of hypothesis to verify the
researchers’ claim.
Research hypothesis:
Rejection region (to reject H0):
Conclusion in terms of the problem:
Possible error in terms of the problem: type 1 error
Error depends on the rejection region
Decision Error type probability
Reject H0 Type 1 error – to reject H0
when it may be true
Alpha – chosen by the
experimenter
Retain H0 Type ii error – to retain Ho
when it may not be true
unknown
b. Compute a 95% CI for the true mean cholesterol level after six months.
c. Repeat parts a and b with n=16
d. How many seniors should be sampled if the true mean cholesterol level should be
estimated within 2 units with 98% confidence. You may use sample standard
deviation of 10 to be an estimate for the population standard deviation.
2. The average birth weight of babies in a local community is 8 lbs. It is hypothesized that
the birth weight of babies born to smoking mothers is less than this value. In a random
sample of 36 babies born to smoking mothers, the mean birth weight was 7.5 lbs with a
standard deviation of 1.5 lbs. With alpha = .05, test the hypothesized claim.
Research hypothesis:
Ha H0:
Test statistic
Rejection region:
Conclusion in terms of the problem:
Possible error in terms of the problem:
b. Construct a 98% confidence interval for the true birth weight of babies born to smoking moms.
c. How many smoking moms should be sampled if the birth weight is to be estimated within half
a pound, with 96% confidence? Assume that the range of birth weight is between 6 to 10 lbs and
range= sigma/4, approximately
3.To test the claim that daily exercise help reduce blood sugar levels of diabetes patients, the
fasting blood sugar levels of 100 diabetes patients after an exercise program were recorded. The
sample average was calculated to be 132 with a standard deviation of 15.2. The sample average
of fasting blood sugar level of 81 patients without any exercise is 140 units with a standard
deviation of 12.
a.The researchers hypothesized that the exercise program should reduce the fasting blood sugar
level.. Assume alpha=.01 and test the researcher’s claim.
b. Construct a 95% confidence interval for the mean fasting blood sugar level of diabetes patients.
c. If the true mean difference of the fasting blood sugar levels between the two groups is to be
estimated within 5 units of the true mean difference with 95% confidence, how many from each
group should be sampled. Assume that the sigma value for both populations are 15 each.
4. When comparing two instructors in public health, the average test scores obtained
by students taught by both teachers were recorded. The sample sizes were 16 for
each group and the sample means were 81 and 79 respectively, while the sample
standard deviations were 5 and 7. Is there sufficient statistically significant
evidence to conclude that teacher1 produces better results than teacher 2.? First
answer the following.
Construct a 99% CI for the true mean difference in test scores
5.In Fall 2016 after a flu epidemic, 5 students out of 100 elementary school students
were found with flu. In fall 2015, the prevalence of flu among elementary school kids
was 20%. Can you conclude that the flu prevalence among elementary school kids
have increased from 2015 to 2016? Justify your answer.
6. The proportion of smokers among college students is known to be about 18%.
25 students out of 100 from a nearby college were found to be smokers. Can
you conclude that there is a higher proportion of smokers in this college
compared to the entire college student community? Justify your answer by
statistical reasoning.