Posted: May 1st, 2022

ceiling. Assume that that distribution of the population is normal.

1. On average, a banana will last 7 days from the time it is purchased in the store to the time it is too rotten to eat. Is the mean time to spoil less if the banana is hung from the ceiling? The data show results of an experiment with 16 bananas that are hung from the ceiling. Assume that that distribution of the population is normal.

7.7, 5.2, 4.5, 6.3, 7.9, 7.7, 8.2, 6.5, 4.2, 6, 5, 6.5, 7.2, 7.6, 4.6, 7.6

What can be concluded at the the αα = 0.01 level of significance level of significance?

For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
The null and alternative hypotheses would be:
H0:H0: ? μ p Select an answer < > = ≠

H1:H1: ? p μ Select an answer = > ≠ <

The test statistic ? z t = (please show your answer to 3 decimal places.)
The p-value = (I need help writing my essay – research paper show your answer to 4 decimal places.)
The p-value is ? ≤ > αα
Based on this, we should Select an answer fail to reject reject accept the null hypothesis.
Thus, the final conclusion is that …
The data suggest the populaton mean is significantly less than 7 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time that it takes for bananas to spoil if they are hung from the ceiling is less than 7.
The data suggest that the population mean time that it takes for bananas to spoil if they are hung from the ceiling is not significantly less than 7 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean time that it takes for bananas to spoil if they are hung from the ceiling is less than 7.
The data suggest the population mean is not significantly less than 7 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean time that it takes for bananas to spoil if they are hung from the ceiling is equal to 7.
2.

It takes an average of 12.5 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will change if the patient is immediately told the truth about the injury. The EMT randomly selected 43 injured patients to immediately tell the truth about the injury and noticed that they averaged 12.3 minutes for their blood to begin clotting after their injury. Their standard deviation was 2.05 minutes. What can be concluded at the the αα = 0.01 level of significance?

a. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion

b.The null and alternative hypotheses would be:

H0:H0: ? μ p Select an answer = ≠ > <

H1:H1: ? p μ Select an answer > ≠ < =

c. The test statistic ? z t = (please show your answer to 3 decimal places.)

d. The p-value = (I need help writing my essay – research paper show your answer to 4 decimal places.)

e. The p-value is ? ≤ > αα

f. Based on this, we should Select an answer accept reject fail to reject the null hypothesis.

g. Thus, the final conclusion is that …

The data suggest the population mean is not significantly different from 12.5 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time for blood to begin clotting after an injury if the patient is told the truth immediately is equal to 12.5.
The data suggest the populaton mean is significantly different from 12.5 at αα = 0.01, so there is statistically significant evidence to conclude that the population mean time for blood to begin clotting after an injury if the patient is told the truth immediately is different from 12.5.
The data suggest that the population mean is not significantly different from 12.5 at αα = 0.01, so there is statistically insignificant evidence to conclude that the population mean time for blood to begin clotting after an injury if the patient is told the truth immediately is different from 12.5.
3.

On average, Americans have lived in 2 places by the time they are 18 years old. Is this average less for college students? The 43 randomly selected college students who answered the survey question had lived in an average of 1.94 places by the time they were 18 years old. The standard deviation for the survey group was 0.4. What can be concluded at the αα = 0.10 level of significance?

a. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion

b. The null and alternative hypotheses would be:

H0:H0: ? p μ Select an answer > ≠ < =

H1:H1: ? p μ Select an answer > < = ≠

c. The test statistic ? t z = (please show your answer to 3 decimal places.)

d. The p-value = (I need help writing my essay – research paper show your answer to 4 decimal places.)

e. The p-value is ? ≤ > αα

f. Based on this, we should Select an answer accept fail to reject reject the null hypothesis.

g. Thus, the final conclusion is that …

The data suggest that the populaton mean is significantly less than 2 at αα = 0.10, so there is statistically significant evidence to conclude that the population mean number of places that college students lived in by the time they were 18 years old is less than 2.
The data suggest that the population mean is not significantly less than 2 at αα = 0.10, so there is statistically insignificant evidence to conclude that the population mean number of places that college students lived in by the time they were 18 years old is less than 2.
The data suggest that the sample mean is not significantly less than 2 at αα = 0.10, so there is statistically insignificant evidence to conclude that the sample mean number of places that college students lived in by the time they were 18 years old is less than 1.94.
h.Interpret the p-value in the context of the study.

If the population mean number of places that college students lived in by the time they were 18 years old is 2 and if you survey another 43 college students, then there would be a 16.54685157% chance that the sample mean for these 43 college students would be less than 1.94.
There is a 16.54685157% chance that the population mean number of places that college students lived in by the time they were 18 years old is less than 2.
If the population mean number of places that college students lived in by the time they were 18 years old is 2 and if you survey another 43 college students, then there would be a 16.54685157% chance that the population mean number of places that college students lived in by the time they were 18 years old would be less than 2.
There is a 16.54685157% chance of a Type I error.
i. Interpret the level of significance in the context of the study.

If the population mean number of places that college students lived in by the time they were 18 years old is 2 and if you survey another 43 college students, then there would be a 10% chance that we would end up falsely concluding that the population mean number of places that college students lived in by the time they were 18 years old is less than 2.
There is a 10% chance that none of this is real since you have been hooked up to virtual reality since you were born.
There is a 10% chance that the population mean number of places that college students lived in by the time they were 18 years old is less than 2.
If the population mean number of places that college students lived in by the time they were 18 years old is less than 2 and if you survey another 43 college students, then there would be a 10% chance that we would end up falsely concluding that the population mean number of places that college students lived in by the time they were 18 years old is equal to 2.

4.

The average number of cavities that thirty-year-old Americans have had in their lifetimes is 4. Do twenty-year-olds have more cavities? The data show the results of a survey of 13 twenty-year-olds who were asked how many cavities they have had. Assume that the distribution of the population is normal.

4, 6, 6, 6, 4, 4, 3, 5, 6, 5, 6, 3, 4

What can be concluded at the αα = 0.01 level of significance?

a. For this study, we should use Select an answer t-test for a population mean z-test for a population proportion

b. The null and alternative hypotheses would be:

H0:H0: ? μ p Select an answer ≠ = > <

H1:H1: ? μ p Select an answer < ≠ > =

c. The test statistic ? z t = (please show your answer to 3 decimal places.)

d. The p-value = (I need help writing my essay – research paper show your answer to 4 decimal places.)

e. The p-value is ? ≤ > αα

f. Based on this, we should Select an answer fail to reject accept reject the null hypothesis.

g. Thus, the final conclusion is that …

The data suggest that the population mean number of cavities for twenty-year-olds is not significantly more than 4 at αα = 0.01, so there is insufficient evidence to conclude that the population mean number of cavities for twenty-year-olds is more than 4.
The data suggest the populaton mean is significantly more than 4 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of cavities for twenty-year-olds is more than 4.
The data suggest the population mean is not significantly more than 4 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of cavities for twenty-year-olds is equal to 4.
h. Interpret the p-value in the context of the study.

There is a 1.74081782% chance of a Type I error.
If the population mean number of cavities for twenty-year-olds is 4 and if you survey another 13 twenty-year-olds then there would be a 1.74081782% chance that the population mean number of cavities for twenty-year-olds would be greater than 4.
If the population mean number of cavities for twenty-year-olds is 4 and if you survey another 13 twenty-year-olds then there would be a 1.74081782% chance that the sample mean for these 13 twenty-year-olds would be greater than 4.77.
There is a 1.74081782% chance that the population mean number of cavities for twenty-year-olds is greater than 4.
i. Interpret the level of significance in the context of the study.

If the population mean number of cavities for twenty-year-olds is more than 4 and if you survey another 13 twenty-year-olds, then there would be a 1% chance that we would end up falsely concuding that the population mean number of cavities for twenty-year-olds is equal to 4.
If the population mean number of cavities for twenty-year-olds is 4 and if you survey another 13 twenty-year-olds, then there would be a 1% chance that we would end up falsely concuding that the population mean number of cavities for twenty-year-olds is more than 4.
There is a 1% chance that the population mean number of cavities for twenty-year-olds is more than 4.
There is a 1% chance that flossing will take care of the problem, so this study is not necessary.

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