Sleep experts believe that sleep apnea is more likely to occur in men than in th
ID: 3365464 • Letter: S
Question
Sleep experts believe that sleep apnea is more likely to occur in men than in the general population, that is the percentage of men who suffer from sleep apnea is greater than 8%. To test this claim, one expert examined 75 men and found that 10 of these men suffer from sleep apnea. What are the Null and Alternative Hypotheses?
Ho: p = 0.08 Ha: p < 0.08
Ho: p 0.08 Ha: p > 0.08
Ho: p = 0.08 Ha: p 0.08
Ho: p > 0.08 Ha: p 0.08
Sleep experts believe that sleep apnea is more likely to occur in men than in the general population, that is the percentage of men who suffer from sleep apnea is greater than 8%. To test this claim, one expert examined 75 men and found that 10 of these men suffer from sleep apnea. Set up the appropriate hypothesis test and find the critical value(s) at the 0.05 Level of significance.
1.645
-1.645
2.33
1.96
Sleep experts believe that sleep apnea is more likely to occur in men than in the general population, that is the percentage of men who suffer from sleep apnea is greater than 8%. To test this claim, one expert examined 75 men and found that 10 of these men suffer from sleep apnea. Set up the appropriate hypothesis test and find the standardized test statistic.
t = 1.70
Z = 1.29
Z = 1.70
p-hat = 0.13
Based on Questions 1 - 3 state a conclusion at the 0.05 Level of Significance.
Reject Ho
Fail to Reject Ho
Accept Ha
Reject Ha
Interpret the results of Question 4.
There is not enough evidence to reject the expert's claim that the percentage of men with sleep apnea is greater than 8%.
There is enough evidence to reject the expert's claim that the percentage of men with sleep apnea is greater than 8%.
The evidence does not support the expert's claim that the percentage of men with sleep apnea is greater than 8%.
The evidence supports the expert's claim that the percentage of men with sleep apnea is greater than 8%.
A farmer decides to try out a new fertilizer on a test plot containing 10 stalks of corn. Before applying the fertilizer, he measures the height of each stalk. Two weeks later, after applying the fertilizer, he measures the stalks again. He compares the heights of these stalks to 10 stalks that did not receive fertilizer. Did the fertilizer help? Use a significance level of 0.10 to test whether the height of the stalks increased.
What are the Null and Alternative Hypotheses?
H0: µd = 0
Ha: µd 0
H0: µd > 0
Ha: µd < 0
H0: µd < 0
Ha: µd > 0
H0: µd = 0
Ha: µd > 0
A farmer decides to try out a new fertilizer on a test plot containing 10 stalks of corn. Before applying the fertilizer, he measures the height of each stalk. Two weeks later, after applying the fertilizer, he measures the stalks again. He compares the heights of these stalks to 10 stalks that did not receive fertilizer. Did the fertilizer help? Use a significance level of 0.10 to test whether the height of the stalks increased.
Set up the appropriate hypothesis test and find the critical value(s) at the 0.10 Level of significance.
t = -1.38
t = -1.83
t = 1.83
z = -1.28
A farmer decides to try out a new fertilizer on a test plot containing 10 stalks of corn. Before applying the fertilizer, he measures the height of each stalk. Two weeks later, after applying the fertilizer, he measures the stalks again. He compares the heights of these stalks to 10 stalks that did not receive fertilizer. Did the fertilizer help? Use a significance level of 0.10 to test whether the height of the stalks increased.
The differences are calculated and the mean difference is found to be -3.36 inches with a standard deviation of 1.05 inches. Set up the appropriate hypothesis test and find the standardized test statistic.
t* = -14.31
t* = 3.2
t* = -3.2
t* = -10.12
Based on Questions 6 - 8 state a conclusion at the 0.05 Level of Significance.
Reject the Null Hypothesis (H0) in favor of the Alternative.
Fail to Reject the Null Hypothesis (H0) in favor of the Alternative.
Reject the Alternative Hypothesis (Ha) in favor of the Null.
Fail to Reject the Alternative Hypothesis (Ha) in favor of the Null.
Interpret the results of Question 9.
There is enough evidence at the alpha = 0.10 level to reject that the fertilizer helps increase the height of the corn stalks.
There is not enough evidence at the alpha = 0.10 level to reject that the fertilizer helps increase the height of the corn stalks.
There is enough evidence at the alpha = 0.10 level to suggest that the fertilizer helps increase the height of the corn stalks.
There is not enough evidence at the alpha = 0.10 level to suggest that the fertilizer helps increase the height of the corn stalks.
A labor union claims that the mean income of its members is no more than $20,000 per year. The Employers' Association believes that this claim is too low and that the actual mean income is higher. A random sample of 64 union members yields a mean income of $20,278.50 with a standard deviation of $793.81. Calculate the test statistic for the sample mean.
H0: µ = 20,000
Ha: µ 20,000
H0: µ < 20,000
Ha: µ > 20,000
H0: µ > 20,000
Ha: µ < 20,000
H0: µ = 20,000
Ha: µ > 20,000
A labor union claims that the mean income of its members is no more than $20,000 per year. The Employers' Association believes that this claim is too low and that the actual mean income is higher. A random sample of 64 union members yields a mean income of $20,278.50 with a standard deviation of $793.81. Find the critical value at the 0.05 level of sinificace.
t = 1.669
t = 1.998
z = 1.645
z = 1.96
A labor union claims that the mean income of its members is no more than $20,000 per year. The Employers' Association believes that this claim is too low and that the actual mean income is higher. A random sample of 64 union members yields a mean income of $20,278.50 with a standard deviation of $793.81. Calculate the standardized test statistic.
-0.35
0.35
-2.81
2.81
A labor union claims that the mean income of its members is no more than $20,000 per year. The Employers' Association believes that this claim is too low and that the actual mean income is higher. A random sample of 64 union members yields a mean income of $20,278.50 with a standard deviation of $793.81. State a conclusion based on this test at the 0.05 level of significance.
Reject Ho
Fail to Reject Ho
Accept Ha
Reject Ha
Interpret the results of Question 14.
There is not enough evidence to reject the Labor Union's claim that the mean salary is $20,000.
There is enough evidence to reject the Labor Union's claim that the mean salary is $20,000.
The evidence supports the Labor Union's claim that the mean salary is $20,000.
The evidence does not support the Labor Union's claim that the mean salary is $20,000.
Ho: p = 0.08 Ha: p < 0.08
Ho: p 0.08 Ha: p > 0.08
Ho: p = 0.08 Ha: p 0.08
Ho: p > 0.08 Ha: p 0.08
Explanation / Answer
Q1 )
c) Ho: p 0.08 Ha: p > 0.08 is correct
Q2)
A) 1.645 is correct
Q3)
Test and CI for One Proportion
Test of p = 0.08 vs p > 0.08
Sample X N Sample p 95% Lower Bound Z-Value P-Value
1 10 75 0.133333 0.068769 1.70 0.044
Using the normal approximation.
C) Z = 1.70 is correct
Q4 )
d) The evidence supports the expert's claim that the percentage of men with sleep apnea is greater than 8%.
is correct
d) The evidence supports the expert's claim that the percentage of men with sleep apnea is greater than 8%.
is correct
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