Download Math 116 - Chapters 14, 15, 17, 18

Math 116 - Chapters 14, 15, 17, 18
On chapter 15: Only sample size will be tested
Name___________________________________
Differentiate between: (a) 1 sample, (b) 2 dependent samples (matched pairs), (3) 2 independent samples
1) A ramdom sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assume
that the population standard deviation is known to be 1.6 cm.
a) Verify assumptions
b) Use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is more than 39.9
cm. First: use the calculator. Second: verify with formulas
c) Construct a 90% confidence interval estimate for the mean circumference of all pumpkins.
First: use the calculator. Second: verify with formulas.
d) What does the interval suggest: mu = 39.9, mu < 39.9 or mu > 39.9? Explain.
e) Explain the meaning of the p-value
f) What do you mean when you say: I am 90% confident that......
g) If we want to produce an estimate that is within .25 cm of the true mean, how many more pumpkins should
be selected?
A) 1 POPULATION MEAN - Z PROCEDURES (with a sigma, use z, chapter 14)
a) The sample is an SRS. The sample size is large, so the x-bar distribution for samples of size 100 is
approximately normal.
b) x-bar = 40.5, sigma = 1.6, n = 100 - We are using a z- test because sigma is given.
H0 : µ = 39.9; H1 : µ > 39.9. Sketch a graph, shade, label center and possible x-bar locations
Test statistic: Z = 3.75. P-value: P (x-bar > 40.5) = 0.00009 < 0.05. This very low p-value gives strong
evidence against the null hypotheis and in favor of the alternative hypothesis. The probability of
observing such a high value of x-bar = 40.5 in a population with a true mean of 39.9 is almost zero (p =
0.00009). This suggests that the true mean of the population is in fact higher than 39.9 cm.
At the 5% significance level there is sufficient evidence to support the claim that the mean circumference
of all pumpkins is more than 39.9 cm. Notice that this conclusion is true at any of the common significance
levels.
c) We are 90% confidence that the mean circumference of all pumpkins is between 40.237 cm and 40.763
cm. Since all values in the interval are more than 39.9, the interval also provides evidence supporting the
claim that the mean circumference of all pumpkins is more than 39.9 cm.
d) The interval provides plausible values for mu. Since all the numbers in the interval are larger than 39.9,
the interval sugests that mu > 39.9
e) p = P(x-bar > 40.5) = 0.00009. It is very unlikely to observe a sample mean of 40.5 cm or more when
selecting a sample of size 100 from a population with mu = 39.9. This is a more likely event in a
population with mu > 39.9.
f) 90% refers to the success rate of the method used. If the process (of selecting 100 pumpkins, finding the
mean of the sample and constructing the corresponding 90% interval) is done MANY times, we expect
about 90% of the intervals constructed to contain mu and about 10% to miss it.
g) Use the formula for sample size given in chapter 15. The sample size n should be 111. Since we have
data from 100 pumpkins we need to select 11 more pumpkins at random and construct a new interval
around the x-bar of the sample of 111 pumpkins. This new interval will have a margin of error of at most
.25.
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2) A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by
following a particular diet. Use the sample data below to test the claim that the treatment population mean µ 1 is
smaller than the control population mean µ2 .
a) Discuss the conditions that should be met in order to proceed with the inference methods learned.
b) What is the sample statistic x-bar1 - x-bar2
c) Test the claim using a significance level of 0.01. Use calculator only.
d) Use the sample data below to construct a 98% confidence interval for u1 - u2 where u1 and u2 represent the
mean for the treatment group and the control group respectively. Use calculator only.
e) What does the interval suggest? u1=u2, u1 < u2, u1>u2. Explain your choice.
f) Use the interval results to find the margin of error
Treatment Group
Control Group
n1 = 85
n 2 = 75
x1 = 189.1
x2 = 203.7
s1 = 38.7
s2 = 39.2
A) 2 INDEPENDENT SAMPLES - Chapter 18 - We will solve this problem with calculator only - no
formulas
a) It should be stated that these are SRS. Since the samples are large, the shape is not a problem because
the Central Limit Theorem applies
b) Our point estimate fo mu(1) - mu(2) is x-bar1 - x-bar2 = - 14.6 (this number should be labeled in
graph)
c) Use a Non-pooled t-procedure.
H0 : µ1 = µ2 or µ1 - µ2=0
(equal means, diet not effective in reducing blood pressure)
H1 : µ1 < µ2 . or µ1 - µ2 < 0 (diet is effective in reducing blood pressure)
Graph, shade, label.
Test statistic t = -2.365. p = 0.0096 < 0.01. P(x-bar1 - x-bar2) < -14.6 = 0.0096. Unlikely that the samples
came from two populations with equal means.
It is not very likely (p = 0.0096) to observe such a difference of sample means (-14.6) when the population
means are equal. At the 1% significance level there is strong evidence supporting the claim that the mean
blood pressure of the diet population µ1 is lower than the mean blood pressure of the control population
µ2 . At the 1% significance level we can say that the diet has worked in reducing blood pressure.
Notice that the conclusion would be different at the significance level of 0.005.
d) -29.11 < µ1 - µ 2 < - 0.089. We are 98% confident that µ1 - µ2 is between -29.11 and -0.089.
e) Since the interval is completely below zero, it suggests that mu1 < mu2 . This agrees with the conclusion
to part (b)
f) ME = -.089 - (-14.6) = 14.511
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3) A cereal company claims that the mean weight of the cereal in its packets is more than 14 oz. The weights (in
ounces) of the cereal in a random sample of 8 of its cereal packets are listed below.
14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2
a) The sample size is small, can you use one of the methods learned in class to perform inference?
b) At the 0.01 significance level, test the claim that the mean weight is more than14 ounces. Use calculator and
then use formulas.
c) Construct a 98% confidence interval estimate for the mean weight of the cereal in the company's packets. Do
the results agree with the hypothesis testing results? Must explain why. Use calculator and then use formulas.
d) We are _____ confident that the mean weight of the cereal packets is _____ with a margin of error of _____
A) 1 SAMPLE T-PROCEDURES - chapter 17
a) Constructing a dot plot for the data shows no extreme outliers or skewness. The robustness of the
t-procedures allow us to perform the tests and obtain approximate results.
b) Ho: mu = 14; Ha: mu > 14. Run a t-test and obtain: Test statistic: t = 0.41. Critical value: t = -2.998.
p-value = .348. (high probability, very likely to observe such a sample mean or a more extreme one when
mu = 14.). The sample mean x-bar is higher than 14 just by chance. There is not sufficient evidence to
support the claim that the mean weight is more than 14 ounces.
c) Run a t-interval and obtain (13.683, 14.417). We are 98% confident that the mean weight of the cereal is
between 13.683 ounces and 14.417 ounces. Since the interval contains 14, it is possible for mu to be 14. This
agrees with the conclusion to part (b)
d) We are 98% confident that the mean weight of the cereal packets is 14.05 oz with a margin of error of
.367
4) A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less
than 10 minutes. Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting
time was 7.8 minutes with a standard deviation of 2.5 minutes.
a) Discuss assumptions of procedures.
b) At the 0.025 significance level, test the claim that the mean is less than 10 minutes.
c) Construct a 95% confidence interval estimate for the mean waiting time of bus number 14. Do the results
agree with the hypothesis testing results? Must explain why
A) 1 SAMPLE T TEST chapter 17
a) We need an SRS and since the sample size is small it is important to know that the shape of this sample
is not strongly skewed and does not contain outliers.
b) Ho: mu = 10, Ha: mu < 10. Test statistic: t = -3.73. p = 0.0008. (x-bar = 7.8 is very unusual when mu =
10)
The probability of observing a sample mean of 7.8 minutes in a population with true mean of 10 minutes
is 0.0008 (almost zero). This suggests that the true mean waiting time for bus number 14 is actually lower
that 10 minutes. There is sufficient evidence to support the claim that the mean waiting time for bus
number 14 during peak hours is less than 10 minutes.
c)With 95% confidence we can say that the mean waiting time for bus number 14 during peak hours is
between 6.5568 minutes and 9.0432 minutes. An interval providea plausible values for mu, since all
numbers in the interval are lower than 10, the interval suggests that mu is lower than 10 which agrees
with the hypohesis testing results.
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5) A coach uses a new technique in training middle distance runners. The times for 8 different athletes to run 800
meters before and after this training are shown below.
Athlete
A
B
C
D
E
F
G
H
Time before
training (seconds) 110.4 117.3 116.1 110.2 114.5 109.8 111.1 112.8
Time after training
(seconds) 111 116 113.7 111 112.7 109.9 107.5 108.9
a) Discuss assumptions
b) Using a 0.05 level of significance, test the claim that the training helps to improve the athletes' times for the
800 meters. Use calculator only.
c) Construct a 90% confidence interval estimate of the mean of the differences between the times before training
and after training. Use calculator only. What does the interval suggest?
A) a) Assuming we have an SRS and there are no strong skewness and outliers we can conduct t-procedures.
into List 3 which is obtained by doing L1 - L2.
b) If we do 1-var Stats of L3 we obtain an d-bar = 1.4375 seconds and s = 1.826. seconds. The finishing
time for the athletes in the sample has dropped an average of 1.4375 seconds. Is d-bar higher than zero by
chance or significantly higher than zero?
H0 : µd = 0, (time before training is equal to time after training which means training has no effect on the
finishing time)
H1: H0 : µ d > 0 (time before training is higher than after training which means that the training improved
the finishing time of the athletes)
Graph, shade, label d-bar in graph.
Test statistic t = 2.227. p-value = 0.031. (Critical value: t = 1.895).
p-value = P(d-bar > 1.4375) = 0.031 < 0.05. It's unusual (p = 0.031) to observe such a d-bar (1.4375) if the
training has no effect on the finishing time. d-bar is significantly higher than zero, which means that the
training lowered the finishing time of the athletes.
At the 5% significance level there is sufficient evidence to support the claim that the training helps to
improve the athletes' times for the 800 meters. Note that if we use a 1% significance level, the conclusion
would be different.
c) (.21441, 2.6606). All numbers in the interval are positive which means that the interval also supports the
claim that ud > 0. with 90% confidence we can say that training helps to improve the athletes' time for the
800 meter. (the mean time before training was higher than the mean time after training).
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Use the computer display to answer the question.
6) A researcher performs a hypothesis test to test the claim that among girls aged 9-10, the mean score on a
particular aptitude test differs from 72, which is the mean score for boys of the same age. The computer display
is shown below. Use the display to identify the following components of the test: the null and alternative
hypotheses, the sample mean, the sample standard deviation, the sample size, the test statistic, and the p-value.
Using a significance level of 0.05, what is your conclusion from these data? Explain how you reached your
conclusion.
TEST OF MU = 72 VS MU
72
N MEAN STDEV SE MEAN T P VALUE
Cl 19 75.88
12.7
2.914
1.33
0.20
H0: µ = 72
A) The hypotheses are:
H1: µ 72
Sample mean is x = 75.88, sample standard deviation is s = 12.7, sample size is n = 19, test statistic is
t = 1.33, p-value is 0.20. Because the p-value of 0.20 is greater than the significance level of 0.05, we do not
reject the null hypothesis. There is not sufficient evidence to accept the claim that the mean score for girls
differs from 72.
7) Jenny is conducting a hypothesis test concerning a population mean. The hypotheses are as follows.
Ho : µ = 50
Ha: µ > 50
She selects a sample and finds that the sample mean is 54.2. She then does some calculations and is able to make
the following statement:
If Ho were true, the chance that the sample mean would have come out as big ( or bigger) than 54.2 is 0.3. Do
you think that she should reject the null hypothesis? Why or why not?
A) Fail to reject Ho. The p-value > any possible alpha. There is a high probability of obtaining an x-bar of
54.2 when mu = 50. It's possible that mu is correct.
B)
8) A one-sample z-test for a population mean is to be performed. Let zo denote the observed value of the test
statistic, z. True or false, for a left-tailed test, the p-value is the area under the standard normal curve to the left
of zo?
A) true
B)
9)
A one-sample z-test for a population mean is performed. Suppose that the p-value for the test is
0.04. For what significance levels (values of ) can the null hypothesis be rejected?
A) any number higher than 0.04
B)
10) A one-sample z-test for a population mean is to be performed. Why might it be more useful for those
interpreting the results to know the p-value rather than simply whether or not the null hypothesis was rejected?
A) If we reject Ho: A small p-value tells you how strong is the evidence against the null and in favor of the
alternative. Also, you will know whether the conclusion depends on the choice of the significance level
alpha or not.
B)
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