MATH 140 -Problem Set 4C
2
Solutions to Practice Problems Section 4C
(#1-10) Use each of the following two-population proportion Z-test statistics and the corresponding critical values to fill out the table.
Z-test
stat
Sentence to explain Z-test statistic.
Critical Value
Does the Z-test statistic fall in a tail determined by a critical value? (Yes or No)
Does sample data significantly disagree with ????0?
4.
-3.177
The sample proportion for group 1 is 3.177 standard errors below the sample proportion for group 2.
±1.96
No
The critical values are ± 1.96. So, the Z-test statistic does not fall in either tail.
Since it does not fall in either tail, the sample data does not significantly disagree with the null hypothesis (????0).
8.
+1.117
The sample proportion for group 1 is 1.117 standard errors above the sample proportion for group 2.
+1.96
Yes
The critical values are +1.96. So, the Z-test statistic falls in the right tail.
Since it falls in one tail, then the sample data significantly disagrees with the null hypothesis.
Table.
(#11-20) Use each of the following P-values and corresponding significance levels to fill out the table.
P-value Proportion
P value %
Sentence to explain the P-value
Significance Level %
Significance level Proportion
If ????0 is true, could the sample data occur by random chance or is it unlikely?
Reject ????0 or Fail to reject ????0?
12.
0.0421
4.21%
The P-value is high above zero (0.0421), indicating a possibility of the sample data occurring because of random chance.
1%
0.01
The sample data could occur by random chance
Fail to reject the null hypothesis
16.
0
0%
The P-value is low (exactly at zero), then it is unlikely to be random chance
10%
0.10
The sample data is unlikely to have occurred by random chance
Reject the null hypothesis
20.
0
0%
The P-value is low (exactly at zero), then it is unlikely to be random chance
5%
0.05
The sample data is unlikely to have occurred by random chance
Reject the null hypothesis
Question 22
The assumptions that we need to check for a two-population proportion hypothesis test
· The sampling method for each population is simple random sampling.
· The samples are independent.
· Each sample includes at least 10 successes and 10 failures.
· Each population is at least 20 times as big as its sample.
Question 26
The United States has the highest teen pregnancy rate in the industrialized world. In 2008, a random sample of 1014 teenage girls found that 326 of them were pregnant before the age of 20. In 2012, a random sample of 1025 teenage girls were taken and 334 were found to be pregnant before the age of 20. Let population proportion 1 represent 2008 and population proportion 2 represent 2012. Use a 10% significance level and the following Statcato printout to test the claim that the population percentage of teen pregnancies in the U.S. is lower in 2008 than it is in 2012. This claim would indicate that the population percentage of U.S. teen pregnancies is related to the year.
Solution
: population percentage of teen pregnancies in the U.S. in 2008
: population percentage of teen pregnancies in the U.S. in 2012
: ? (the population percentage of U.S. teen pregnancies is not related to the year.)
: = (the population percentage of U.S. teen pregnancies is related to the year.) (claim)
Assumptions
Both samples were collected randomly, Data values within and between the samples should
be independent of each other. At least 10 success and at least 10 failures.
Z-test statistic = -0.210
The sample proportion for group 1 was 0.210 standard errors bellow the sample proportion for group 2
The sample proportions are significantly different since the test statistic fell in the tail determined by the critical value
P-value = 0.4168 or 41.68%
If the null hypothesis is true, then there is about a 41.68% probability of getting the sample data or more extreme because of sampling variability.
The P-value is close to zero, so it is very unlikely that the sample data occurred because of sampling variability.
Since the P-value is greater than the significance level, we fail to reject the null hypothesis.
There is significant evidence to support the claim that (the population percentage of U.S. teen pregnancies is related to the year.
Question 29
A health magazine claims that marriage status is one of the most telling factors for a persons happiness. Use a 10% significance level and the Statcato printout below to test the claim that the percent of married people that are unhappy is lower than the percent of single or divorced people that are unhappy. If this is the case, then perhaps being married, single or divorced is related to being unhappy. The following sample data was collected randomly. Population 1 represented married adults and population 2 represented single or divorced adults.
Solution
: Population proportion of married people that are unhappy
Population proportion of non-married people that are unhappy
: ? ( (Being married is not related to being unhappy)
: = (Being married is related to being unhappy) (claim)
Assumptions
Both samples were collected randomly, Data values within and between the samples are independent of each other. At least 10 success and at least 10 failures
Z-test statistic = ?2.325
The sample proportion for group 1 was 2.325 standard errors bellow the sample proportion for group 2
The sample proportions are significantly different since the test statistic fell in the tail
determined by the critical value
P-value = 0.0100 = 1.0%
If the null hypothesis is true, then there is about a 1.0% probability of getting the sample data or more extreme because of sampling variability.
The P-value is less than the significance level, so it is unlikely that the sample data occurred
because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that married people have a lower percentage of unhappiness than non-married people. This also indicates that being married or not is related to being unhappy.
Question 31
A body mass index of 20-25 indicates that a person is of normal weight for their height and body type. A random sample of 760 women found that 198 of the women had a normal BMI. A random sample of 745 men found that 273 of them had a normal BMI. A fitness magazine claims that the percent of women with a normal BMI is lower than the percent of men with a normal BMI. This would imply that gender is related to having a normal BMI. Let population 1 be the proportion of women with a normal BMI and population 2 be the proportion of men with a normal BMI. Use a 10% significance level and a randomized simulation in StatKey.
Solution
: Population proportion of women with a normal BMI
: Population proportion of men with a normal BMI
: ? (Gender is NOT related to having a normal BMI)
: = (Gender is related to having a normal BMI) (claim)
The sample proportion difference is -0.109
The original sample difference of -0.109 does fall in the tail determined by the significance level and the simulation. This implies that the sample proportion for women was significantly lower than the sample proportion for men.
P-value = 0 = 0%
If the null hypothesis is true, there is a 0% probability of getting this sample data or more extreme because of sampling variability.
It is very unlikely that this sample data occurred because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that percentage of women with normal BMI is lower than the percentage of men. This also implies that having a normal BMI is related to gender.
Question 34
A study was done to see if there is a relationship between the age of a person (teen or adult) and using text messages to communicate. A random sample of 800 teens (population 1) found that 696 of them use text messages regularly to communicate. A random sample of 2252 adults (population 2) found that 1621 of them use text messages regularly to communicate. Test the claim that population percentages are equal for the two groups implying that age is not related to using text messages. Use a randomized simulation in StatKey, and a 10% significance level to test this claim.
Solution
: Population proportion of teens who use text message
: Population proportion of adults who use text message
: ? (age is related to using text messages)
: = (age is not related to using text messages) (claim)
The sample proportion difference is 0.150
The original sample difference of ? 0.098 does fall in the tail determined by the significance
level and the simulation. This implies that the sample proportion for smokers was significantly lower than the sample proportion for non-smokers.
P-value = 0 = 0%
If the null hypothesis is true, there is a 0% probability of getting this sample data or more
extreme because of sampling variability
It is very unlikely that this sample data occurred because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that age is not related to using text messages
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Home>Mathematics homework help>Statistics homework help>MATH 140 -Problem Set 4C
2
Solutions to Practice Problems Section 4C
(#1-10) Use each of the following two-population proportion Z-test statistics and the corresponding critical values to fill out the table.
Z-test
stat
Sentence to explain Z-test statistic.
Critical Value
Does the Z-test statistic fall in a tail determined by a critical value? (Yes or No)
Does sample data significantly disagree with ????0?
4.
-3.177
The sample proportion for group 1 is 3.177 standard errors below the sample proportion for group 2.
±1.96
No
The critical values are ± 1.96. So, the Z-test statistic does not fall in either tail.
Since it does not fall in either tail, the sample data does not significantly disagree with the null hypothesis (????0).
8.
+1.117
The sample proportion for group 1 is 1.117 standard errors above the sample proportion for group 2.
+1.96
Yes
The critical values are +1.96. So, the Z-test statistic falls in the right tail.
Since it falls in one tail, then the sample data significantly disagrees with the null hypothesis.
Table.
(#11-20) Use each of the following P-values and corresponding significance levels to fill out the table.
P-value Proportion
P value %
Sentence to explain the P-value
Significance Level %
Significance level Proportion
If ????0 is true, could the sample data occur by random chance or is it unlikely?
Reject ????0 or Fail to reject ????0?
12.
0.0421
4.21%
The P-value is high above zero (0.0421), indicating a possibility of the sample data occurring because of random chance.
1%
0.01
The sample data could occur by random chance
Fail to reject the null hypothesis
16.
0
0%
The P-value is low (exactly at zero), then it is unlikely to be random chance
10%
0.10
The sample data is unlikely to have occurred by random chance
Reject the null hypothesis
20.
0
0%
The P-value is low (exactly at zero), then it is unlikely to be random chance
5%
0.05
The sample data is unlikely to have occurred by random chance
Reject the null hypothesis
Question 22
The assumptions that we need to check for a two-population proportion hypothesis test
· The sampling method for each population is simple random sampling.
· The samples are independent.
· Each sample includes at least 10 successes and 10 failures.
· Each population is at least 20 times as big as its sample.
Question 26
The United States has the highest teen pregnancy rate in the industrialized world. In 2008, a random sample of 1014 teenage girls found that 326 of them were pregnant before the age of 20. In 2012, a random sample of 1025 teenage girls were taken and 334 were found to be pregnant before the age of 20. Let population proportion 1 represent 2008 and population proportion 2 represent 2012. Use a 10% significance level and the following Statcato printout to test the claim that the population percentage of teen pregnancies in the U.S. is lower in 2008 than it is in 2012. This claim would indicate that the population percentage of U.S. teen pregnancies is related to the year.
Solution
: population percentage of teen pregnancies in the U.S. in 2008
: population percentage of teen pregnancies in the U.S. in 2012
: ? (the population percentage of U.S. teen pregnancies is not related to the year.)
: = (the population percentage of U.S. teen pregnancies is related to the year.) (claim)
Assumptions
Both samples were collected randomly, Data values within and between the samples should
be independent of each other. At least 10 success and at least 10 failures.
Z-test statistic = -0.210
The sample proportion for group 1 was 0.210 standard errors bellow the sample proportion for group 2
The sample proportions are significantly different since the test statistic fell in the tail determined by the critical value
P-value = 0.4168 or 41.68%
If the null hypothesis is true, then there is about a 41.68% probability of getting the sample data or more extreme because of sampling variability.
The P-value is close to zero, so it is very unlikely that the sample data occurred because of sampling variability.
Since the P-value is greater than the significance level, we fail to reject the null hypothesis.
There is significant evidence to support the claim that (the population percentage of U.S. teen pregnancies is related to the year.
Question 29
A health magazine claims that marriage status is one of the most telling factors for a persons happiness. Use a 10% significance level and the Statcato printout below to test the claim that the percent of married people that are unhappy is lower than the percent of single or divorced people that are unhappy. If this is the case, then perhaps being married, single or divorced is related to being unhappy. The following sample data was collected randomly. Population 1 represented married adults and population 2 represented single or divorced adults.
Solution
: Population proportion of married people that are unhappy
Population proportion of non-married people that are unhappy
: ? ( (Being married is not related to being unhappy)
: = (Being married is related to being unhappy) (claim)
Assumptions
Both samples were collected randomly, Data values within and between the samples are independent of each other. At least 10 success and at least 10 failures
Z-test statistic = ?2.325
The sample proportion for group 1 was 2.325 standard errors bellow the sample proportion for group 2
The sample proportions are significantly different since the test statistic fell in the tail
determined by the critical value
P-value = 0.0100 = 1.0%
If the null hypothesis is true, then there is about a 1.0% probability of getting the sample data or more extreme because of sampling variability.
The P-value is less than the significance level, so it is unlikely that the sample data occurred
because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that married people have a lower percentage of unhappiness than non-married people. This also indicates that being married or not is related to being unhappy.
Question 31
A body mass index of 20-25 indicates that a person is of normal weight for their height and body type. A random sample of 760 women found that 198 of the women had a normal BMI. A random sample of 745 men found that 273 of them had a normal BMI. A fitness magazine claims that the percent of women with a normal BMI is lower than the percent of men with a normal BMI. This would imply that gender is related to having a normal BMI. Let population 1 be the proportion of women with a normal BMI and population 2 be the proportion of men with a normal BMI. Use a 10% significance level and a randomized simulation in StatKey.
Solution
: Population proportion of women with a normal BMI
: Population proportion of men with a normal BMI
: ? (Gender is NOT related to having a normal BMI)
: = (Gender is related to having a normal BMI) (claim)
The sample proportion difference is -0.109
The original sample difference of -0.109 does fall in the tail determined by the significance level and the simulation. This implies that the sample proportion for women was significantly lower than the sample proportion for men.
P-value = 0 = 0%
If the null hypothesis is true, there is a 0% probability of getting this sample data or more extreme because of sampling variability.
It is very unlikely that this sample data occurred because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that percentage of women with normal BMI is lower than the percentage of men. This also implies that having a normal BMI is related to gender.
Question 34
A study was done to see if there is a relationship between the age of a person (teen or adult) and using text messages to communicate. A random sample of 800 teens (population 1) found that 696 of them use text messages regularly to communicate. A random sample of 2252 adults (population 2) found that 1621 of them use text messages regularly to communicate. Test the claim that population percentages are equal for the two groups implying that age is not related to using text messages. Use a randomized simulation in StatKey, and a 10% significance level to test this claim.
Solution
: Population proportion of teens who use text message
: Population proportion of adults who use text message
: ? (age is related to using text messages)
: = (age is not related to using text messages) (claim)
The sample proportion difference is 0.150
The original sample difference of ? 0.098 does fall in the tail determined by the significance
level and the simulation. This implies that the sample proportion for smokers was significantly lower than the sample proportion for non-smokers.
P-value = 0 = 0%
If the null hypothesis is true, there is a 0% probability of getting this sample data or more
extreme because of sampling variability
It is very unlikely that this sample data occurred because of sampling variability.
Since the P-value is less than the significance level, we will reject the null hypothesis.
There is significant evidence to support the claim that age is not related to using text messages
Applied Sciences
Architecture and Design
Biology
Business & Finance
Chemistry
Computer Science
Geography
Geology
Education
Engineering
English
Environmental science
Spanish
Government
History
Human Resource Management
Information Systems
Law
Literature
Mathematics
Nursing
Physics
Political Science
Psychology
Reading
Science
Social Science
Liberty University
New Hampshire University
Strayer University
University Of Phoenix
Walden University
Home
Homework Answers
Blog
Archive
Tags
Reviews
Contact
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