Download StpT3Key - Arizona State University

Exam #3A Key
1. You want to estimate the mean weight of all students at the large
university. A random sample of n = 10 students gives a sample mean
of 145 lb. and a sample standard deviation of 15 lb. Assume that
the weights of the students at that university are normally
distributed.
a. (8points) Obtain 95% confidence interval for , the true mean
weight. Keep 2 decimal spaces in your computations.
( t interval, since  unknown, df=9)
(145- 2.262*15/10, 145+2.262*15/10)
gives (134.27, 155.73)
b. (8points) Looking at your interval, do you think that there is
evidence at 5% significance level that true mean may be
 = 143 lb? Clearly explain your answer.
Yes, we have 95% confidence that our CI captures the true mean and
143 is in the interval.
c) (8 points) Suppose that you know that the population standard
deviation is 14 lb. What should be the sample size so that the 95%
confidence interval for  will have a margin of error of 4 lb?
n=(1.96*14/4)2=48
2. Consider the following:
You are testing the hypotheses:
H0 :  = 1.8
Ha :  > 1.8
You are using  = 0.05.
Your test is based on a random sample of size n = 42. The
population standard deviation  is known.
The observed test statistic z = 2.34.
Use the above information to answer the following questions:
 (8points) What is the rejection region for that test? Clearly mark
critical value(s).
Right sided rejection region (Z  1.645)
 (8points) What is the p- value?
p-value=P(z  2.34) =.0096
 (8points) Do you reject H0 or not? Why or why not? Explain what
your decision means for the alternative hypothesis tested.
Reject H0, since our test statistics falls into the rejection region (and pvalue  .05). There is evidence at 5% significance level for Ha , i.e.
evidence that  > 1.8.
3. The blood cholesterol values for two independent randomly selected
samples of patients were compared for two diets, one low fat and
other normal. The sample means, standard deviations and sizes are
as follows:
Low fat
Normal
x 1  170
x 2  196
s2  15.86
s1  14.07
n1=16
n2=14
 Do the data provide sufficient evidence to indicate that real mean
cholesterol level of patients on low fat diet is smaller than the real
mean cholesterol level of patients on normal diet? Test appropriate
hypothesis using  = 0.05.
Assume that populations from which the samples were taken are
normally distributed and have equal standard deviations.
a. (7points) State null and alternative hypotheses, using appropriate
symbols.
H0 : 1 = 2
Ha : 1< 2
(1= true mean colesterol level of patients on low fat diet
2= true mean colesterol level of patients on normal diet)
b. (7points) Compute appropriate test statistics:
Use pooled t test.
Sp=14.93, t= -4.76
c. (7points) Rejection region (or p-value)
Left sided rejection region (t  -1.701)
P-value=P(t  -4.76) < .005
d. (7 points) Do you reject null hypothesis or not. Explain your decision
and answer the question posed in the problem.
Reject H0, because p-value < .05 (and test statistics falls into rejection
region). There is evidence at %5 significance level for alternative
hypothesis. Patients on low fat diet appear to have lower cholesterol
level than the ones on normal diet.
4. During a 1998 race for state senator a newspaper conducted a poll
and found that 602 of 1200 registered voters sampled would vote for the
Republican candidate. Let p be the population proportion of registered
voters who would vote for the Republican.
a. (8 points) Give a 90 % level confidence interval for p.
(Keep 3 decimal places in your computations)
602/1200=.502
(.502- 1.645*(.502*.498/1200), .502+1.645*(.502*.498/1200))
gives (.478, .526)
b. (8 points) Using your answer to part a), would you conclude that a
Republican is now likely to win? Assume that he needs
at least 50% of votes to win. Explain.
We have 90% confidence that true fraction of voters that would vote
for
a Republican is in our CI. Since it contains both fractions above and
below .5, it is not clear that a Republican will win. Farther more poling
is like taking a temperature of voters at a particular moment in time.
Their opinion may change at the time of voting.
c. (8 points) To obtain a margin of error of 0.01 in your 90% CI for p,
how large must the sample be?
Use .5 or .052 as your best guess for p.
n=.25(1.645/.01)2=6766