1. 200 Hamburgers, Mean = 8, SD = .54. Buy 5 random hamburgers -Mean = 7.6. What is the population and the parameter?

Population: all 200 hamburger pattiesParameter: μ = 8 oz.

9. Interpret the standard deviation.

If many samples they would typically vary by 4.02 and mean of 18.

2. 200 Hamburgers, Mean = 8, SD = .54. Buy 5 random hamburgers --> Mean = 7.6. What is the sample and the statistic?

Sample: 5 hamburger patties ordered Statistic: ̅ 7.6 oz.

4. Is there convincing evidence that the mean weight of hamburgers at this restaurant is less than 8 oz?

Yes, it is possible that the true mean is less then μ = 8 oz.

3. Would it be unusual to get a sample mean of 7.6 oz or less in a sample of size n = 5 from this population? Explain.

Yes. Only 1 sample had a mean weight of 7.6 oz. or less.

5. Based on the distribution of the three statistics, which is preferred as an estimate of the parameter?

Statistic A. It is unbiased and is the least variable.

10. Justify that the distribution of X is approximately normal.

np = 18 > 10 & n(1-p) = 162 > 10

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Chapter 6 Review 1

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1
1. 200 Hamburgers, Mean = 8, SD = .54. Buy 5 random hamburgers -Mean = 7.6. What is the population and the parameter?
Population: all 200 hamburger pattiesParameter: μ = 8 oz.
Population: all 400 hamburger patties Parameter: μ = 8 oz.
Population: all 150 hamburger patties Parameter: μ = 4 oz.
Population: all 120 hamburger patties Parameter: μ = 1 oz.
2
6. List all 6 possible SRSs of size n = 2, and calculate the mean midterm score for each sample.
88,85,81,79,75,72
98,85,81,79,75,72
88
88,85,81
3
9. Interpret the standard deviation.
If many samples they would typically vary by 1.02 and mean of 12.
If many samples they would typically vary by 2.02 and mean of 15.
If many samples they would typically vary by 3.02 and mean of 10.
If many samples they would typically vary by 4.02 and mean of 18.
4
2. 200 Hamburgers, Mean = 8, SD = .54. Buy 5 random hamburgers --> Mean = 7.6. What is the sample and the statistic?
Sample: 4 hamburger patties ordered Statistic: ̅ 8.6 oz.
Sample: 3 hamburger patties ordered Statistic: ̅ 7.6 oz.
Sample: 4 hamburger patties ordered Statistic: ̅ 7.6 oz.
Sample: 5 hamburger patties ordered Statistic: ̅ 7.6 oz.
5
4. Is there convincing evidence that the mean weight of hamburgers at this restaurant is less than 8 oz?
Yes, it is possible that the true mean is less then μ = 5 oz.
Yes, it is possible that the true mean is less then μ = 6 oz.
No, it is possible that the true mean is less then μ = 8 oz.
Yes, it is possible that the true mean is less then μ = 8 oz.
6
7. How would the sampling distribution change if we took SRSs of size n = 3 instead?
The variability will stop.
The variability will decrease.
The variability will increase.
The variability will remain constant.
7
Calculate the probability that 26 or more individuals in the sample do not wear their seat belt.
z=5.99 - (1-.9767) = .8321
z=1.99 - (1-.9767) = .0233
z=-1.99 - (1-.9767) = .9233
z=.5 - (1-.9767) = .1543
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3. Would it be unusual to get a sample mean of 7.6 oz or less in a sample of size n = 5 from this population? Explain.
Yes. Only 1 sample had a mean weight of 2.6 oz. or less.
Yes. Only 1 sample had a mean weight of 7.6 oz. or less.
No. Only 1 sample had a mean weight of 7.6 oz. or less.
Yes. Only 1 sample had a mean weight of 8.6 oz. or less.
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5. Based on the distribution of the three statistics, which is preferred as an estimate of the parameter?
Statistic A. It is unbiased and is the least variable.
Statistic B. It is unbiased and is the least variable.
Statistic D. It is unbiased and is the least variable.
Statistic C. It is unbiased and is the least variable.
10
8. 10% not wear a seatbelt. 180 U.S. residents. Calculate the mean and standard deviation?
Mean = 10, SD = .02
Mean = 18, SD = 4.02
Mean = 12, SD = 1.02
Mean = 17, SD = 3.02
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10. Justify that the distribution of X is approximately normal.
np = 1 > 10 & n(1-p) = 1 > 10
np = 18 > 10 & n(1-p) = 162 > 10
np = 18 < 10 & n(1-p) = 162 < 10
np = 7 > 10 & n(1-p) = 2 > 10

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