QMB 3200 Exam 3
A neighborhood bottler in Hawaii needs to make certain that a mean of 20 ounces of ardour
fruit juice is used to fill each bottle. In order to investigate the accuracy of the bottling
method, she takes a random pattern of 37 bottles. The imply weight of the ardour fruit juice
within the pattern is 19.40 ounces. Assume that the population standard deviation is 1.34
ounce.
A. Specify the null and the opportunity hypotheses to test whether or not the bottling manner
is wrong.
B-1. Calculate the cost of the check statistic.
B-2. Find the p-fee.
C-1. What is the realization at α = zero.10?
C-2. Make a advice to the bottler. - ANS-a. H0: µ = 20; as opposed to HA: µ ≠ 20
b-1. Test statistic: -2.72
b-2. P-price < 0.01
In Excel: =2*NORM.DIST(-2.7236, 0, 1, TRUE) = 0.0065.
C-1. Reject H0 since the value of the p-value is less than the significance level.
C-2. The accuracy of the bottling process is compromised
A newly hired basketball coach promised a high-paced attack that will put more points on the
board than the team's previously tepid offense historically managed. After a few months, the
team owner looks at the data to test the coach's claim. He takes a sample of 49 of the
team's games under the new coach and finds that they scored an average of 105 points with
a standard deviation of 8 points. Over the past 10 years, the team had averaged 103 points.
What is the value of the appropriate test statistic to test the new coach's claim at the 1%
significance level? - ANS-t48 = 1.75
A parameter is a random variable, whereas a sample statistic is a constant. - ANS-False
A promising start-up wants to compete in the cell phone market. It understands that the lead
product has a battery life of approximately 12 hours. The start-up claims that while its new
cell phone is more expensive, its battery life is more than twice as long as that of the leading
product. In order to test the claim, a researcher samples 45 units of the new cell phone and
finds that the sample battery life averages 25.3 hours with a sample standard deviation of
3.1 hours.
A. Specify the relevant null and the alternative hypotheses.
B-1. Calculate the value of the test statistic.
B-2. Find the p-value.
C. At the 5% significance level, what is the conclusion? - ANS-a. H0:µ ≤ 24; versus HA:µ >
24
b-1. Test statistic: 2.813
, b-2. P-cost < zero.01
c. Reject H0; the declare is supported via the records.
A random sample of 15 observations is used to estimate the populace imply. The sample
suggest and the sample preferred deviation are calculated as 152.Five and 28.90,
respectively. Assume that the population is generally allotted.
A. Construct the 95% confidence c programming language for the populace suggest.
B. Construct the 99% self assurance interval for the populace imply.
C. Use your answers to talk about the effect of the confidence degree on the width of the c
language. - ANS-a. 136.50 to 168.50
In Excel:
Lower Limit: =152.Five-CONFIDENCE.T(0.05,28.Nine,15)
Upper Limit: =152.Five+CONFIDENCE.T(zero.05,28.Nine,15)
b. One hundred thirty.29 to 174.Seventy one
In Excel:
Lower Limit: =152.5-CONFIDENCE.T(0.01,28.Nine,15)
Upper Limit: =152.5+CONFIDENCE.T(zero.01,28.9,15)
c. As the confidence degree will increase, the c language turns into wider.
A researcher desires to determine if Americans are sleeping much less than the advocated 7
hours of sleep on weekdays. He takes a random sample of one hundred Americans and
computes the common sleep time of 6.7 hours on weekdays. Assume that the populace is
usually allotted with a regarded standard deviation of 2.1 hours. Test the researcher's
declare at α = zero.01.
A. Specify the null and the alternative hypotheses for the take a look at.
B. Calculate the fee of the test statistic.
C. Find the p-value.
D. What is the conclusion at α = zero.01?
E. Make an inference. - ANS-a. H0:µ ≥ 7; versus HA:µ < 7
b. Test statistic: -1.Forty three
c. 0.05 ≤ p-price < 0.10
d. Do no longer reject H0 since the p-cost is extra than α.
E. There is insufficient proof to indicate that Americans sleep less than the recommended 7
hours of sleep.
A pattern of 47 observations yields a pattern mean of 59.7. Assume that the pattern is drawn
from a ordinary population with a population preferred deviation of 5.7.
A-1. Find the p-value.
A-2. What is the belief if α = zero.10?
A-3. Interpret the consequences at α = zero.10.
B-1. Calculate the p-cost if the above pattern suggest changed into based on a pattern of
126 observations.
B-2. What is the belief if α = zero.10?
B-3. Interpret the outcomes at α = zero.10. - ANS-a-1. P-price ≥ 0.10
a-2. Do not reject H0 because the p-cost is extra than α.
A-three. We can not conclude that the populace mean is extra than 58.7.
B-1. Zero.01 ≤ p-cost < 0.1/2
b-2. Reject H0 for the reason that p-value is smaller than α.
B-three. We finish that the population imply is more than 58.7.
A neighborhood bottler in Hawaii needs to make certain that a mean of 20 ounces of ardour
fruit juice is used to fill each bottle. In order to investigate the accuracy of the bottling
method, she takes a random pattern of 37 bottles. The imply weight of the ardour fruit juice
within the pattern is 19.40 ounces. Assume that the population standard deviation is 1.34
ounce.
A. Specify the null and the opportunity hypotheses to test whether or not the bottling manner
is wrong.
B-1. Calculate the cost of the check statistic.
B-2. Find the p-fee.
C-1. What is the realization at α = zero.10?
C-2. Make a advice to the bottler. - ANS-a. H0: µ = 20; as opposed to HA: µ ≠ 20
b-1. Test statistic: -2.72
b-2. P-price < 0.01
In Excel: =2*NORM.DIST(-2.7236, 0, 1, TRUE) = 0.0065.
C-1. Reject H0 since the value of the p-value is less than the significance level.
C-2. The accuracy of the bottling process is compromised
A newly hired basketball coach promised a high-paced attack that will put more points on the
board than the team's previously tepid offense historically managed. After a few months, the
team owner looks at the data to test the coach's claim. He takes a sample of 49 of the
team's games under the new coach and finds that they scored an average of 105 points with
a standard deviation of 8 points. Over the past 10 years, the team had averaged 103 points.
What is the value of the appropriate test statistic to test the new coach's claim at the 1%
significance level? - ANS-t48 = 1.75
A parameter is a random variable, whereas a sample statistic is a constant. - ANS-False
A promising start-up wants to compete in the cell phone market. It understands that the lead
product has a battery life of approximately 12 hours. The start-up claims that while its new
cell phone is more expensive, its battery life is more than twice as long as that of the leading
product. In order to test the claim, a researcher samples 45 units of the new cell phone and
finds that the sample battery life averages 25.3 hours with a sample standard deviation of
3.1 hours.
A. Specify the relevant null and the alternative hypotheses.
B-1. Calculate the value of the test statistic.
B-2. Find the p-value.
C. At the 5% significance level, what is the conclusion? - ANS-a. H0:µ ≤ 24; versus HA:µ >
24
b-1. Test statistic: 2.813
, b-2. P-cost < zero.01
c. Reject H0; the declare is supported via the records.
A random sample of 15 observations is used to estimate the populace imply. The sample
suggest and the sample preferred deviation are calculated as 152.Five and 28.90,
respectively. Assume that the population is generally allotted.
A. Construct the 95% confidence c programming language for the populace suggest.
B. Construct the 99% self assurance interval for the populace imply.
C. Use your answers to talk about the effect of the confidence degree on the width of the c
language. - ANS-a. 136.50 to 168.50
In Excel:
Lower Limit: =152.Five-CONFIDENCE.T(0.05,28.Nine,15)
Upper Limit: =152.Five+CONFIDENCE.T(zero.05,28.Nine,15)
b. One hundred thirty.29 to 174.Seventy one
In Excel:
Lower Limit: =152.5-CONFIDENCE.T(0.01,28.Nine,15)
Upper Limit: =152.5+CONFIDENCE.T(zero.01,28.9,15)
c. As the confidence degree will increase, the c language turns into wider.
A researcher desires to determine if Americans are sleeping much less than the advocated 7
hours of sleep on weekdays. He takes a random sample of one hundred Americans and
computes the common sleep time of 6.7 hours on weekdays. Assume that the populace is
usually allotted with a regarded standard deviation of 2.1 hours. Test the researcher's
declare at α = zero.01.
A. Specify the null and the alternative hypotheses for the take a look at.
B. Calculate the fee of the test statistic.
C. Find the p-value.
D. What is the conclusion at α = zero.01?
E. Make an inference. - ANS-a. H0:µ ≥ 7; versus HA:µ < 7
b. Test statistic: -1.Forty three
c. 0.05 ≤ p-price < 0.10
d. Do no longer reject H0 since the p-cost is extra than α.
E. There is insufficient proof to indicate that Americans sleep less than the recommended 7
hours of sleep.
A pattern of 47 observations yields a pattern mean of 59.7. Assume that the pattern is drawn
from a ordinary population with a population preferred deviation of 5.7.
A-1. Find the p-value.
A-2. What is the belief if α = zero.10?
A-3. Interpret the consequences at α = zero.10.
B-1. Calculate the p-cost if the above pattern suggest changed into based on a pattern of
126 observations.
B-2. What is the belief if α = zero.10?
B-3. Interpret the outcomes at α = zero.10. - ANS-a-1. P-price ≥ 0.10
a-2. Do not reject H0 because the p-cost is extra than α.
A-three. We can not conclude that the populace mean is extra than 58.7.
B-1. Zero.01 ≤ p-cost < 0.1/2
b-2. Reject H0 for the reason that p-value is smaller than α.
B-three. We finish that the population imply is more than 58.7.
