Unless stated otherwise/For all applicable questions:
For each claim/question, state the:
(A.) Null and alternative hypothesis in words.
(B.) Null and alternative hypothesis in symbols.
(C.) Parameter that is being tested.
(D.) Type of hypothesis test.
(E.) Other relevant information as applicable.
NOTE:
If the question deals with proportion and you are not given a population proportion, use 50% or 0.5 as the population proportion as the null hypothesis.
Then, calculate the sample proportion, compare it with the population and use that comparison (less than, greater than, or not equal to; as applicable) to determine the alternative hypothesis.
The names of the towns (in italics) are written to make you smile.
Yes, those towns exist. 😊
I want to make this topic fun as much as I can.
(1.) A recent poll estimated that 4.5% of American adults are vegetarian.
A nutritionist thinks this rate has increased and will take a random sample of American adults and record whether or not they are vegetarian.
$
4.5\% = \dfrac{4.5}{100} = 0.045 \\[5ex]
$
(A.)
Hint for the null hypothesis: 4.5% of American adults are vegetarian Null Hypothesis: The proportion of American adults who are vegetarian is 4.5%
Hint for the alternative hypothesis: A nutritionist thinks this rate has increased. Alternative Hypothesis: The proportion of American adults who are vegetarian is greater than 4.5%
(B.)
$
H_0:\:\: p = 0.045 \\[3ex]
H_1:\:\: p \gt 0.045 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
(2.) Three years ago, the mean price of a single-family home in the village of Loving, New Mexico (would they surpass the State of Virginia for lovers?) was
$100,000
A realtor believes that the mean price has increased since then.
(A.) Null Hypothesis: The average price of a single-family home in the village is $100,000 Alternative Hypothesis: The average price of a single-family home in the village is greater than $100,000
(3.) The standard deviation of pulse rates of adult males in the town of Goat Town, Georgia (do they have a lot of goats?) is more than 12
bpm (beats per minute).
For a random sample of 129 adult males, the pulse rates have a standard deviation of 12.3 bpm.
(A.) Null Hypothesis: The standard deviation of the pulse rates of adult males in the town is $12$ bpm Alternative Hypothesis: The standard deviation of the pulse rates of adult males in the town is more than $12$ bpm
(4.) The mean pulse of adult males in the town of Coward, South Carolina (oh wow...are they cowards?) is 69.3 bpm (beats per minute).
For a random sample of 162 adult means, the mean pulse rate is 69.5 bpm, and the standard deviation is 11.4 bpm
(A.) Null Hypothesis: The mean pulse rate of adults in the town is $69.3$ bpm Alternative Hypothesis: The mean pulse rate of adults in the town is not $69.3$
(B.)
$
H_0:\:\: \mu = 69.3 \\[3ex]
H_1:\:\: \mu \ne 69.3 \\[3ex]
$
(C.) One-sample mean
(D.) Two-tailed test
(5.) In the City of Surprise, Arizona (be prepared for surprises. Do you like surprises?), the birth weights of 32 babies whose mothers do not smoke and the birth weights
of 28 babies whose mothers smoke, were compared.
Test the hypothesis that the population mean birth weight is larger for mothers who do not smoke.
Assume the sample is random, and the distributions are normal.
Use a significance level of 0.05.
(A.)
Let:
$\mu_1$ be the mean weight of babies of mothers who do not smoke.
$\mu_2$ be the mean weight of babies of mothers who smoke.
Null Hypothesis: The mean birth weight of the babies of mothers who do NOT smoke is the same as the mean birth weight of
the babies of mothers who smoke. Alternative Hypothesis: The mean birth weight of the babies of mothers who do NOT smoke is greater than the mean birth weight of
the babies of mothers who smoke.
(6.) Investigators performed a randomized experiment in which 418 juvenile delinquents in the town of Rough and Ready, California (must you get rough before you get ready?)
were randomly assigned to either multisystemic therapy (MST) or just probation (control group).
Of the 215 assigned to therapy, 92 had criminal convictions within 12 months.
Of the 203 in the control group, 74 had criminal convictions within 12 months.
Determine whether the therapy caused significantly fewer arrests using a significance level of 1%
(A.)
Let:
$p_1$ be the proportion of arrests for delinquents assigned to MST.
$p_2$ be the proportion of arrests for delinquents assigned to probation.
Null Hypothesis: The proportion of arrests for delinquents who received multisystemic therapy is the same as the proportion
of arrests for delinquents who received probation. Alternative Hypothesis: The proportion of arrests for delinquents who received multisystemic therapy is less than the proportion
of arrests for delinquents who received probation.
(7.) Veronica was asked to differentiate distilled water from tap water.
Fifty trials (half with distilled water, and half with tap water) was given.
She got forty-five trials correctly.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
Use 50% for p because it was not given
$
p = 50\% = \dfrac{50}{100} = 0.5 \\[5ex]
0.9 \gt 0.5 \\[3ex]
$
(A.) Null Hypothesis: The population proportion is 0.5 Alternative Hypothesis: The population proportion is greater than 0.5
(B.)
$
H_0:\:\: p = 0.5 \\[3ex]
H_1:\:\: p \gt 0.5 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
(8.) Most adults in the community of Happyland, Oklahoma (happiest people in the USA?) would not erase all their tweets online permanently if they could.
A survey of 430 randomly selected adults by Concise Polls showed that 44% of them would erase all their tweets
online permanently if they could.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
$
\hat{p} = 44\% = \dfrac{44}{100} = 0.44 \\[5ex]
Use\:\:50\%\:\:for\:\:p\:\:because\:\:it\:\:was\:\:not\:\:given \\[3ex]
p = 50\% = \dfrac{50}{100} = 0.5 \\[5ex]
0.44 \lt 0.5 \\[3ex]
$
(A.) Null Hypothesis: The population proportion is $0.5$ Alternative Hypothesis: The population proportion is less than $0.5$
(B.)
$
H_0:\:\: p = 0.5 \\[3ex]
H_1:\:\: p \lt 0.5 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Left-tailed test)
(9.) A fair coin is flipped fifty times and lands on heads forty times.
Gideon wants to test the hypothesis that the coin does not come up seventy percent heads in the long run.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
$
\hat{p} = 70\% = \dfrac{70}{100} = 0.7 \\[5ex]
However,\:\:we\:\:shall\:\:use\:\:it\:\:as\:\:the\:\:population\:\:proportion \\[3ex]
Does\:\:not\:\:come\:\:up\:\:seventy\:\:percent\:\:heads\:\:means \ne 0.7 \\[3ex]
$
(A.) Null Hypothesis: The population proportion is $0.7$ Alternative Hypothesis: The population proportion is NOT equal to $0.7$
(B.)
$
H_0:\:\: p = 0.7 \\[3ex]
H_1:\:\: p \ne 0.7 \\[3ex]
$
(C.) One-sample proportion
(D.) Two-tailed test
(10.) A fair die is flipped thirty-five times and lands on a 1, five times.
Joshua wants to test the hypothesis that the die does not come up with a 1, one-sixth of the time in the long run.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
$
\hat{p} = \dfrac{1}{6} \\[5ex]
$
However, we shall use it as the population proportion
Does not come up with 1 means ≠ $\dfrac{1}{6}$
$
(A.) Null Hypothesis: The population proportion is one-sixth Alternative Hypothesis: The population proportion is NOT one-sixth
(B.)
$
H_0:\:\: p = \dfrac{1}{6} \\[5ex]
H_1:\:\: p \ne \dfrac{1}{6} \\[5ex]
$
(C.) One-sample proportion
(D.) Two-tailed test
(11.) An experiment was done with criminals released from prison in the town of Neversink, New York (do not sink, there is still hope to rise)
who attend boot camp before being released from prison.
The recidivism rate was found to be 24%.
This means that 24% of the criminals released from prison return to prison within three years.
One hundred random prisoners are made to attend a boot camp for two weeks prior to their release, with the expectation
that the boot camp will have a good effect to reduce the recidivism rate.
$24\% = \dfrac{24}{100} = 0.24$
(A.) Null Hypothesis: The proportion of criminals who attend boot campy and return to prison within three years is $0.24 (24\%)$ Alternative Hypothesis: The proportion of criminals who attend boot campy and return to prison within three years is less than $0.24 (24\%)$
('less than' because of 'reduce the recidivism rate')
(B.)
$
H_0:\:\: p = 0.24 \\[3ex]
H_1:\:\: p \lt 0.24 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Left-tailed test)
(12.) A college in the City of Los Baños, California (how frequent do they use the bathrooms?) wants to increase
its retention rate (the number of new students who begin one year and return the next year) of forty five percent.
After implementing several new programs to improve retention over a two-year period, the college randomly sampled
three hundred and sixty students who started one year and found that one hundred and ninety four of them returned the
next year.
The college wants to determine if the initiative is working to increase retention.
Remember: we are concerned with the population parameter.
$45\% = \dfrac{45}{100} = 0.45$
(A.) Null Hypothesis: The retention rate is $0.45 (45\%)$ Alternative Hypothesis: The retention rate is more than $0.45 (45\%)$
('more than' because of 'wants to increase its retention rate')
(B.)
$
H_0:\:\: p = 0.45 \\[3ex]
H_1:\:\: p \gt 0.45 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
(13.) A 95% confidence interval for the ages of six consecutive presidents at their inaugurations is about (50.9, 62.4).
Either interpret the interval or explain why it should not be interpreted.
A. It should not be interpreted.
The data is not Normal and so inference based on a confidence interval is not possible.
B. We are 95% confident that the mean of all president's ages is between 50.9 and 62.4.
C. We are 95% confident that the mean of all president's ages is not between 50.9 and 62.4.
D. It should not be interpreted.
The data are not a random sample and so inference based on a confidence interval is not possible.
D. It should not be interpreted.
The data are not a random sample and so inference based on a confidence interval is not possible.
The data was collected systematically, selecting 6 consecutive presidents.
Since it was not collected randomly, this is not a valid confidence interval.
(14.) A formal hypothesis test is to be conducted to test the claim that the wait times at the Space Mountain ride in Walt Disney World have a mean equal to 46 minutes.
(E.) What are the possible conclusions that can be made about the null hypothesis?
(F.) What are the possible conclusions that can be made about the alternative hypothesis?
(G.) Is it possible to conclude that there is sufficient evidence to support the claim that the mean wait time is equal to minutes?
(i.) Yes
(ii.) No. Hypothesis tests can never find sufficient evidence to support a claim; they can only find sufficient evidence to warrant rejection of a claim.
(iii.) No. Hypothesis tests can only find sufficient evidence to support a claim when the claim does not contain equality.
(iv.) No. Hypothesis tests can only find sufficient evidence to support a claim when the claim contains equality.
(A.) Null Hypothesis: The population mean is 46 minutes Alternative Hypothesis: The population mean is not equal to 46 minutes
(B.)
$
H_0:\:\: \mu = 46\;minutes \\[3ex]
H_1:\:\: \mu \ne 46\;minutes \\[3ex]
$
(C.) One-sample mean
(D.) Two-tailed test
(E.) The possible conclusions that can be made about the null hypothesis are:
(a.) Fail to reject the null hypothesis.
(b.) Reject the null hypothesis.
(E.) The possible conclusions that can be made about the alternative hypothesis are:
(a.) Fail to reject the alternative hypothesis.
(b.) Reject the alternative hypothesis.
(G.) No. Hypothesis tests can only find sufficient evidence to support a claim when the claim does not contain equality.
(15.) A college chemistry instructor thinks the use of embedded tutors (tutors who work with students during regular class meeting times) will improve the success rate in introductory chemistry courses.
The passing rate for introductory chemistry is 64%.
The instructor will use embedded tutors in all sections of introductory chemistry and record the percentage of students passing the course.
$
\hat{p} = \dfrac{35}{40} = 0.875 \\[5ex]
0.875 \gt 0.5 \\[3ex]
$
(A.) Null Hypothesis: The proportion of students who pass introductory chemistry is equal to 64% for classes with embedded tutors. Alternative Hypothesis: The proportion of students who pass introductory chemistry is greater than 64% for classes with embedded tutors.
(B.)
$
H_0:\:\: p = 0.64 \\[3ex]
H_1:\:\: p \gt 0.64 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
(16.) Claim: The standard deviation of pulse rates of adult males is less than 12bpm.
For a random sample of 152 adult males, the pulse rates have a standard deviation of 11.7bpm.
(a.) Parameter: Population Standard Deviation
Original Claim: $\sigma \lt 12$bpm
(17.) The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl.
In a clinical trial, results could be analyzed with a formal hypothesis test with the alternative hypothesis of p > 0.5, which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5.
If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, 0.001? Why?
One-sample proportion
$
H_0:\;\; p = 0.5 \\[3ex]
H_1:\;\; p \gt 0.5 \\[3ex]
$
A smaller p-value indicates a stronger evidence in favor of the alternative hypothesis.
This implies that the P-value of 0.001 is most preferred.
The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the hypothesis that the method is effective.
(18.) Recently, a health organization estimated that the flu vaccine was 69% effective against the influenza B virus.
An immunologist suspects that the current flu vaccine is less effective against this virus.
$
69\% = \dfrac{69}{100} = 0.69 \\[5ex]
$
(A.)
A health organization estimated that the flu vaccine was 69% effective against the influenza B virus Null Hypothesis: The proportion is 0.69
An immunologist suspects that the current flu vaccine is less effective against this virus. Alternative Hypothesis: The proportion is less than 0.69
(B.)
$
H_0:\:\: p = 0.69 \\[3ex]
H_1:\:\: p \lt 0.69 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Left-tailed test)
(19.) A local news agency in Pee Pee Township, Ohio (do they have to pee all the time!?) reports that
forty seven percent of the adult population regularly uses supplemental vitamins.
A manager at a local grocery store believes that number is significantly different and collects data from one hundred and
seventy randomly selected adults from the city to test the claim.
The table summarizes the data.
Adults Surveyed
Regularly takes supplemental vitamins
91
Does not regularly takes supplemental vitamins
79
Total
170
$47\% = \dfrac{47}{100} = 0.47$
(A.) Null Hypothesis: The proportion of the population that regularly uses supplemental vitamins is $0.47 (47\%)$ Alternative Hypothesis: The proportion of the population that regularly uses supplemental vitamins is NOT equal to $0.47 (47\%)$
('NOT equal to' because of 'significantly different')
(B.)
$
H_0:\:\: p = 0.47 \\[3ex]
H_1:\:\: p \ne 0.47 \\[3ex]
$
(C.) One-sample proportion
(even though the table makes it seems as if it is two samples: the sample that regularly uses supplemental vitamins and the sample that does not regularly use supplemental vitamins. Review the wording of the question.)
(D.) Two-tailed test
(this is because of 'believes that number is significantly different')
(20.) A researcher made a claim that the proportion of urban families who regularly eat lunch together is statistically
lower than the proportion of rural families.
The table displays the summary of her data collection survey.
$x_1$ is the number of urban families who regularly eat lunch.
$n_1$ is the total number of urban families.
$x_2$ is the number of rural families who regularly eat lunch.
$n_2$ is the total number of rural families.
Urban Families Surveyed
Rural Families Surveyed
$x_1$ = 12
$x_2$ = 40
$n_1$ = 75
$n_2$ = 140
Keep in mind that the null hypothesis and alternative hypothesis deals with population parameters.
$p_1$ is the first population proportion (population of urban families who regularly eat lunch together)
$p_2$ is the second population proportion (population of rural families who regularly eat lunch together)
(A.) Null Hypothesis: The proportion of urban families who regularly eat lunch together is equal to the proportion of rural families who regularly eat lunch together Alternative Hypothesis: The proportion of urban families who regularly eat lunch together is lower than the proportion of rural families who regularly eat lunch together
(B.)
$
H_0:\:\: p_1 = p_2 \\[3ex]
H_1:\:\: p_1 \lt p_2 \\[3ex]
$
(C.) Two-sample proportion
The two samples are: the sample of urban families who regularly eat lunch together and the sample of rural who regularly eat lunch together
(21.) Claim: More than 3.3% of homes have only a landline telephone and no wireless phone.
Sample data: A survey by the National Center for Health Statistics showed that among 14961 homes, 5.76% had landline phones without wireless phones.
(a.) Express the original claim in symbolic form.
Let the parameter represent a value with respect to homes that have only a landline telephone and no wireless phone.
(b.)
$
H_0:\:\: p = 0.033 \\[3ex]
H_1:\:\: p \gt 0.033 \\[3ex]
$
(c.) One-sample proportion
(d.) One-tailed test (Right-tailed test)
(22.) According to a recent poll at a university, 71.2% of high school seniors in a certain region had a driver's license.
A sociologist thinks this rate has declined.
The sociologist surveys 500 randomly selected high school seniors and finds that 350 have a driver's license.
$
71.2\% = \dfrac{71.2}{100} = 0.712 \\[5ex]
$
(A.)
71.2% of high school seniors in a certain region had a driver's license. Null Hypothesis: The proportion of high school seniors in the entire region who has a driver's license is 0.712
A sociologist thinks this rate has declined. Alternative Hypothesis: The proportion of high school seniors in the entire region who has a driver's license is less than 0.712
(B.)
$
H_0:\:\: p = 0.712 \\[3ex]
H_1:\:\: p \lt 0.712 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Left-tailed test)
(23.) A law association estimated that 85.8% of law school graduates last year found employment.
An economist thinks the current employment rate for law school graduates is different from the rate last year.
$
85.8\% = \dfrac{85.8}{100} = 0.858 \\[5ex]
$
(A.)
A law association estimated that 85.8% of law school graduates last year found employment. Null Hypothesis: The proportion of law school graduates last year that found employment is 0.858
An economist thinks the current employment rate for law school graduates is different from the rate last year. Alternative Hypothesis: The proportion of law school graduates last year that found employment is not 0.858
(B.)
$
H_0:\:\: p = 0.858 \\[3ex]
H_1:\:\: p \ne 0.858 \\[3ex]
$
(C.) One-sample proportion
(D.) Two-tailed test
(24.) Claim: The mean systolic blood pressure of all healthy adults is less than 121mmHg.
Sample data: For 292 healthy adults, the mean systolic blood pressure level is 119.27mmHg and the standard deviation is 16.18mmHg.
(a.) Parameter: Population Mean
Original Claim: $\mu \lt 121$mmHg
(25.) The manufacturer of a certain engine treatment in the City of Truth or Consequences, New Mexico claims that if you add their product to your engine, it will be protected from excessive wear.
An infomercial claims that a woman drove 6 hours without oil, thanks to the engine treatment.
A magazine tested engines in which they added the treatment to the motor oil, ran the engines, drained the oil, and then determined the time until the engines seized.
Write the null and alternative hypothesis.
Both engines took exactly 17 minutes to seize.
What conclusion might the magazine make based on this evidence?
$
H_0:\;\; \mu = 6 \\[3ex]
H_1:\;\; \mu \lt 6 \\[3ex]
$
The infomercial’s claim is not true.
(26.) A friend is tested to see whether he can tell bottled water from tap water.
There are 40 trials (half with bottled water and half with tap water), and he gets 25 right.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
$
\hat{p} = \dfrac{25}{40} = 0.625 \\[5ex]
$
Use 50% for p because it was not given
$
p = 50\% = \dfrac{50}{100} = 0.5 \\[5ex]
0.625 \gt 0.5 \\[3ex]
$
(A.) Null Hypothesis: The population proportion is 0.5 Alternative Hypothesis: The population proportion is greater than 0.5
(B.)
$
H_0:\:\: p = 0.5 \\[3ex]
H_1:\:\: p \gt 0.5 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
(27.)
(28.) The label on a can of mixed nuts says that the mixture contains 50% peanuts.
After opening a can of nuts and finding 27 peanuts in a can of 50 nuts, a consumer thinks the proportion of peanuts in the mixture differs from 50%.
The consumer writes these hypotheses
$
H_0:\;\; p = 0.50 \\[3ex]
H_1:\;\; p = 0.54 \\[3ex]
$
where p represents the proportion of peanuts in all cans of mixed nuts from this company.
Are these hypotheses written correctly?
Correct any mistakes as needed.
A. The hypotheses are not correct.
They should be H0: p = 0.50 and H1: p > 0.50
B. The hypotheses are not correct.
They should be H0: p = 0.50 and H1: p ≠ 0.50
C. The hypotheses are not correct.
They should be H0: p = 0.54 and H1: p ≠ 0.54
D. The hypotheses are not correct.
They should be H0: p = 0.54 and H1: p > 0.54
E. The hypotheses are not correct.
They should be H0: p = 0.54 and H1: p < 0.54
F. The hypotheses are not correct.
They should be H0: p = 0.50 and H1: p < 0.50
G. The hypotheses are correct.
$
50\% = \dfrac{50}{100} = 0.50 \\[5ex]
\dfrac{27}{50} = 0.54 \\[5ex]
$
The label on a can of mixed nuts says that the mixture contains 50% peanuts. Null Hypothesis: The proportion of peanuts in a can of mixed nuts is 0.50
H0: p = 0.50
After opening a can of nuts and finding 27 peanuts in a can of 50 nuts, a consumer thinks the proportion of peanuts in the mixture differs from 50%. BUT: from the question:
Alternative Hypothesis: The proportion of peanuts in a can of mixed nuts is greater than 0.50
H1: p ≠ 0.50
The answer is Option B.
(29.) A manager at a casual dining restaurant noted that 14% of customers ordered soda with their meal.
In an effort to increase soda sales, the restaurant begins offering free refills with every soda order for a two-week trial period.
During this trial period, 18% of customers ordered soda with their meal.
To test if the promotion was successful in increasing soda orders, the manager wrote the following hypotheses
$
H_0:\;\; p = 0.14 \\[3ex]
H_1:\;\; \hat{p} = 0.18 \\[3ex]
$
where $\hat{p}$ represents the proportion of customers who ordered soda with their meal during promotion.
Are these hypotheses written correctly?
Correct any mistakes as needed.
A. The hypotheses are not correct.
They should be H0: p = 0.14 and H1: p < 0.14
B. The hypotheses are not correct.
They should be H0: p = 0.18 and H1: p < 0.18
C. The hypotheses are not correct.
They should be H0: p = 0.14 and H1: p ≠ 0.14
D. The hypotheses are not correct.
They should be H0: p = 0.18 and H1: p > 0.18
E. The hypotheses are not correct.
They should be H0: p = 0.14 and H1: p > 0.14
F. The hypotheses are not correct.
They should be H0: p = 0.18 and H1: p ≠ 0.18
G. The hypotheses are correct.
$
14\% = \dfrac{14}{100} = 0.14 \\[5ex]
18\% = \dfrac{18}{100} = 0.18 \\[5ex]
$
...14% of customers ordered soda with their meal. Null Hypothesis: The proportion of customers who ordered soda with their meal is 0.14
H0: p = 0.14
During this trial period, 18% of customers ordered soda with their meal.
0.18 > 0.14 Alternative Hypothesis: The proportion of customers who ordered soda with their meal is greater than 0.14
H1: p > 0.14
The answer is Option E.
(30.) A 2003 study of dreaming found that out of a random sample of 114 people, 82 reported dreaming in color.
However, the rate of reported dreaming in color that was established in the 1940s was 0.21.
Check to see whether the conditions for using a one-proportion z-test are met assuming the researcher wanted to test to see if the proportion dreaming in color had changed since the 1940s.
Condition 1: Random Sample: ...that out of a random sample of 114 people,...
Condition 1 is met.
Condition 2 is also met.
$
\underline{Condition\;3} \\[3ex]
x = 82 \\[3ex]
n = 114 \\[3ex]
p = 0.21 \\[3ex]
q = 1 - 0.21 = 0.79 \\[3ex]
np = 114(0.21) = 23.94 \gt 10 \\[3ex]
nq = 114(0.79) = 90.06 \gt 10 \\[3ex]
Condition\;3\;\;is\;\;met \\[3ex]
$
The conditions for using a one-proportion z-test are met.
(31.) The table below lists leading digits of 317 inter-arrival Internet traffic times for a computer, along with the frequencies of leading digits expected with Benford's law.
When using these data to test for goodness-of-fit with the distribution described by Benford's law, identify the:
(a.) Null hypotheses
(b.) Alternative hypotheses.
Leading Digit
1
2
3
4
5
6
7
8
9
Benford's Law
30.1%
17.6%
12.5%
9.7%
7.9%
6.7%
5.8%
5.1%
4.6%
Leading Digits of Inter-Arrival Traffic Times
76
62
29
33
19
27
28
21
22
H0: p1 = 0.301 and p2 = 0.176 and p3 = 0.125 ...and... p9 = 0.046
H1: At least one of the proportions is not equal to the given claimed value.
(32.)
(33.)
(34.) A coin is flipped 40 times and 35 heads come up.
The claim is for the coin to come up heads more than 50 of the time.
Even though this case is a sample proportion, our hypothesis test should focus on the population proportion.
We were not given the population proportion, so we shall use 0.5 as the population proportion.
$
\hat{p} = \dfrac{35}{40} = 0.875 \\[5ex]
0.875 \gt 0.5 \\[3ex]
$
(A.) Null Hypothesis: The population proportion is 0.5 Alternative Hypothesis: The population proportion is greater than 0.5
(B.)
$
H_0:\:\: p = 0.5 \\[3ex]
H_1:\:\: p \gt 0.5 \\[3ex]
$
(C.) One-sample proportion
(D.) One-tailed test (Right-tailed test)
Unless stated otherwise/For all applicable questions: