Introduction to Hypothesis Testing (T)
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Transcript of Introduction to Hypothesis Testing (T)
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7/27/2019 Introduction to Hypothesis Testing (T)
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PR2103
Pharmacostatistics
2013
Dr Perry Lim
Introduction toHypothesis Testing
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Q1: The mean lung capacity for non-smokers was reported to be
2 L. A random sample of 40 smokers has a mean lung capacity
of 1.8 L. is assumed to be 0.35 L. Is the lung capacity ofsmokers different from that of non-smokers at a significance level
of 5%?
35.0;2;8.1;40 0 xn ( is known)z-score
There is no difference between lung capacities of smokers and
non-smokers.
n
xz
0 61.3
4035.0
28.1
z
Lung capacity of smokers is different from that of non-smokers.
and are sample and population mean lung capacities of smokers.
0 is lung capacity of non-smokers.
x
H0: = 2 L
H1: 2 L
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Q1: The mean lung capacity for non-smokers was reported to be
2 L. A random sample of 40 smokers has a mean lung capacity
of 1.8 L. is assumed to be 0.35 L. Is the lung capacity ofsmokers different from that of non-smokers at a significance level
of 5%?
5% significance level; = 0.05
z/2 = z0.025 = 1.96
z = |3.61| > |1.96| RejectH0
Conclusion: At a significance level of 0.05, there is a difference
between the lung capacities of smokers and non-smokers.
Acceptanceregion
z1.961.96
Rejection region
3.61
Rejection region
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Q2:A consumer group is investigating a producer of diet meals
to examine if their pre-packaged meals do contain the advertised
protein amount of 6 oz in each package. Based on the followingdata, is there any evidence that the meals do not contain the
advertised protein amount at a significance level of 1%?
6.16.05.85.65.54.94.24.74.85.2
5.86.06.14.95.55.75.16.04.95.1
n
xx
1
)( 2
n
xxs
395.5x 552.0s
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Q2:A consumer group is investigating a producer of diet meals
to examine if their pre-packaged meals do contain the advertised
protein amount of 6 oz in each package. Based on the followingdata, is there any evidence that the meals do not contain the
advertised protein amount at a significance level of 1%?
Pre-packaged diet meals contain advertised protein amount.
90.4
20552.0
6395.5
t
552.0;6;395.5;20 0 sxn ( is not known) t-score
ns
xt 0
Pre-packaged diet meals do not contain advertised protein amount.
and are sample and population mean protein amounts in diet meals.0 is advertised protein amount in diet meals.x
H0: = 6 oz
H1: 6 oz
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Q2:A consumer group is investigating a producer of diet meals
to examine if their pre-packaged meals do contain the advertised
protein amount of 6 oz in each package. Based on the followingdata, is there any evidence that the meals do not contain the
advertised protein amount at a significance level of 1%?
1% significance level; = 0.01 df= n 1 = 19
t/2,df= t0.005,19 = 2.861
Acceptanceregion
Rejection region
t2.8612.861
Rejection region
4.90
t = |4.90| > |2.861| RejectH0
Conclusion: At a significance level of 0.01, the pre-packaged diet
meals do not contain the advertised amount of protein.
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Q3: The mean hospital stay following a particular surgical
procedure in 2007 was reported to be 7.1 days. A researcher
feels that the mean hospital stay in 2008 could be shorter. Arandom sample of 45 patients undergoing the same surgical
procedure in 2008 had a mean stay of 6.24 days with a standard
deviation of 3.01 days. Run the appropriate test at = 0.05.
01.3;1.7;24.6;45 0 sxn ( is not known) t-score
Mean hospital stay in 2008 is not shorter than that in 2007.
ns
x
t0
Mean hospital stay in 2008 is shorter than that in 2007.
92.14501.3
1.724.6
t
and are sample and population mean hospital stays in 2008.0 is mean hospital stay in 2007.x
H0: = 7.1 days
H1: < 7.1 days
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Q3: The mean hospital stay following a particular surgical
procedure in 2007 was reported to be 7.1 days. A researcher
feels that the mean hospital stay in 2008 could be shorter. Arandom sample of 45 patients undergoing the same surgical
procedure in 2008 had a mean stay of 6.24 days with a standard
deviation of 3.01 days. Run the appropriate test at = 0.05.
5% significance level; = 0.05
Acceptance
region
t1.681
Rejection region
1.92
t = |1.92| > |1.681| RejectH0
Conclusion: At a significance level of 0.05, mean hospital stay in
2008 is shorter than that in 2007.
df= n 1 = 44
t,df= t0.05,44 = 1.681
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Q4:A companys cookie recipe states the sugar coating on a
cookie to be 16.4 g. A sample of 28 cookies taken at random has
a mean sugar coating weight of 16.6 g with standard deviation of1.2 g. Is the sugar coated on the cookies more than the stated
amount at 95% confidence level?
Stated amount of sugar is coated on the cookies.
88.0
282.1
4.166.16
t
2.1;4.16;6.16;28 0
sxn ( is not known)
t-score
ns
xt 0
Sugar coated on the cookies is more than stated amount.
and are sample and population mean sugar coating weights.
0 is stated sugar coating weight.
x
H0: = 16.4 g
H1: > 16.4 g
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Q4:A companys cookie recipe states the sugar coating on a
cookie to be 16.4 g. A sample of 28 cookies taken at random has
a mean sugar coating weight of 16.6 g with standard deviation of1.2 g. Is the sugar coated on the cookies more than the stated
amount at 95% confidence level?
df= n 1 = 27
t,df= t0.05,27 = 1.703
Acceptanceregion
t1.703
Rejection region
0.88
t = 0.88 < 1.703 Fail to reject H0
Conclusion: At 95% confidence level, the stated amount of sugar is
coated on the cookies.
95% confidence level; = 0.05
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Q5:A certain supplier has always supplied goods that is 6%
defective. A random sample of 250 units of the goods supplied
recently is found to be 3% defective. Is there an improvement inthe quality of the goods supplied at 95% confidence level?
06.0;03.0;250 0 ppn
There is no improvement in quality of goods supplied.
npp
ppz
)1(
00
0
There is an improvement in quality of goods supplied.
00.2250)06.01(06.0
06.003.0
z
andp are sample and population defective proportions of recent
goods, andp0
is defective proportion of previous goods.
p
H0:p = 0.06
H1:p < 0.06
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Q5:A certain supplier has always supplied goods that is 6%
defective. A random sample of 250 units of the goods supplied
recently is found to be 3% defective. Is there an improvement inthe quality of the goods supplied at 95% confidence level?
z =
z0.05 = 1.645
Acceptance
region
z1.645
Rejection region
2.00
z = |2.00| > |1.645| Reject H0
Conclusion: At 95% confidence level, there is an improvement in
the quality of the goods supplied.
95% confidence level; = 0.05