1. A study of 40 people found that they could do on the average15 pull ups with a standard deviation of .6. Find the 99% confidence interval for the mean of the population.
2. A study of 50 pizza delivery workers found that they could make 6 deliveries per hour with a standard deviation of 1. Find the 95% confidence interval of the mean for all pizza delivery workers.
3. A random sample showed that the average number of tv's in each household is 2.3 with a standard deviation of .4. Find the 90% confidence level for the average number of tv's in every household in the population.
4. A researcher revealed that the average number of people who have a driver's license out of a sample of 100 people is 1.5 with a standard deviation of .3. Find the 80% confidence interval for the mean of the population.
5. The following random sample was selected : 4, 6, 3, 5, 9, 3. Find the 95% confidence interval for the mean of the population.
6. In a sample of 35 high school seniors, 14 of them are attending college in the fall. Find the 95% confidence interval for the true proportion of high school seniors that will attend college in the fall from the population.
7. In a sample of 200 people, 76 people would rather work out at home than in a gym. Find the 99% confidence interval for the true proportion of people who would rather work out at home than in a gym for the entire population.
8. A study found that out of 300 people 60% of them prefer to eat hamburgers rather than hot dogs. Fin the 95% confidence interval for the true proportion of people who prefer to eat hamburgers rather than hot dogs in the entire population.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
Again, I will start you out. We use SEm instead of SD, because we are dealing with distributions of means rather than raw scores.
1. 99% = mean ± 2.575 SEm
SEm = SD/√n
The table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005) to get Z score
(score in standard deviations from the mean).
1. A study of 40 people found that they could do on the average15 pull ups with a standard deviation of .6. Find the 99% confidence interval for the mean of the population.
Confidence Interval:
±0.24
Range for the true population mean:
14.76 to 15.24
2. A study of 50 pizza delivery workers found that they could make 6 deliveries per hour with a standard deviation of 1. Find the 95% confidence interval of the mean for all pizza delivery workers.
Confidence Interval:
±0.28
Range for the true population mean:
5.72 to 6.28
3. A random sample of 50 households showed that the average number of tv's in each household is 2.3 with a standard deviation of .4. Find the 90% confidence level for the average number of tv's in every household in the population.
Confidence Interval:
±0.09
Range for the true population mean:
2.21 to 2.39
4. A researcher revealed that the average number of people who have a driver's license out of a sample of 100 people is 1.5 with a standard deviation of .3. Find the 80% confidence interval for the mean of the population.
Confidence Interval:
±0.04
Range for the true population mean:
1.46 to 1.54
5. The following random sample was selected : 4, 6, 3, 5, 9, 3. Find the 95% confidence interval for the mean of the population.
Confidence Interval:
±0.24
Range for the true population mean:
1.76 to 2.24
6. In a sample of 35 high school seniors, 14 of them are attending college in the fall. Find the 95% confidence interval for the true proportion of high school seniors that will attend college in the fall from the population.
0.14 +/- (1.96)(√0.14)/(0.86)/35)
0.16436428065
0.1643 - .5/1000 = 0.1638 -.0005 = 0.1633
0.1643 + .5/1000 = 0.1648 + .0005 = 0.1653
16.33%-16.53%
7. In a sample of 200 people, 76 people would rather work out at home than in a gym. Find the 99% confidence interval for the true proportion of people who would rather work out at home than in a gym for the entire population.
0.76 +/- (2.575) ((√0.76)/(-0.24)/200)
0.71323264675
0.713 - .5/1000 = 0.7125 -.0005 = 0.712
0.713 + .5/1000 = 0.7135 + .0005 = 0.714
71.2%-71.4%
To find the confidence interval for the mean or proportion of a population, we can use the formula:
Confidence Interval = Sample Statistic ± Margin of Error
The Margin of Error depends on the level of confidence and the sample size. Here are the steps to find the confidence intervals for the given scenarios:
1. For the 99% confidence interval for the mean:
Given: Sample mean (x̄) = 15, Standard deviation (σ) = 0.6, Sample size (n) = 40
First, calculate the standard error (SE) using the formula:
SE = σ / √n
SE = 0.6 / √40 = 0.0949
Next, calculate the margin of error (ME) using the formula:
ME = Critical value * SE
Since the level of confidence is 99%, we need to find the critical value associated with that level. Assuming a normal distribution, the critical value (z*) can be found using a z-table or calculator. For a 99% confidence level, the critical value is approximately 2.576.
ME = 2.576 * 0.0949 = 0.2443
Finally, calculate the confidence interval:
CI = x̄ ± ME
CI = 15 ± 0.2443 = (14.7557, 15.2443)
Therefore, the 99% confidence interval for the average number of pull-ups in the entire population is (14.7557, 15.2443).
2. For the 95% confidence interval for the mean:
Given: Sample mean (x̄) = 6, Standard deviation (σ) = 1, Sample size (n) = 50
Following the same steps as above, calculate the standard error (SE):
SE = 1 / √50 = 0.1414
Find the critical value (z*) corresponding to a 95% confidence level, which is approximately 1.96.
Calculate the margin of error (ME):
ME = 1.96 * 0.1414 = 0.2774
Construct the confidence interval:
CI = x̄ ± ME
CI = 6 ± 0.2774 = (5.7226, 6.2774)
Thus, the 95% confidence interval for the mean number of deliveries per hour for all pizza delivery workers is (5.7226, 6.2774).
3. For the 90% confidence interval for the mean:
Given: Sample mean (x̄) = 2.3, Standard deviation (σ) = 0.4
As there is no sample size mentioned, we assume a large sample or a population.
Since we don't have the sample size, we cannot calculate the standard error. However, we can use the critical value (z*) associated with a 90% confidence level, which is approximately 1.645 for a large sample.
Calculate the margin of error (ME):
ME = z* * (σ / √n)
Since we don't have the sample size (n), we cannot calculate the margin of error or construct the confidence interval.
4. For the 80% confidence interval for the mean:
Given: Sample mean (x̄) = 1.5, Standard deviation (σ) = 0.3, Sample size (n) = 100
Calculate the standard error (SE):
SE = 0.3 / √100 = 0.03
Find the critical value (z*) for an 80% confidence level, which is approximately 1.282.
Calculate the margin of error (ME):
ME = 1.282 * 0.03 = 0.0385
Construct the confidence interval:
CI = x̄ ± ME
CI = 1.5 ± 0.0385 = (1.4615, 1.5385)
Hence, the 80% confidence interval for the mean number of people who have a driver's license in the population is (1.4615, 1.5385).
5. For the 95% confidence interval for the mean:
Given: Sample data: 4, 6, 3, 5, 9, 3
Calculate the sample mean (x̄) and the sample standard deviation (s) using the given data.
x̄ = (4 + 6 + 3 + 5 + 9 + 3) / 6 = 5
s = √((4-5)² + (6-5)² + (3-5)² + (5-5)² + (9-5)² + (3-5)²) / √6 ≈ 2.16
Since the sample size (n) is relatively small (less than 30) and the population standard deviation is unknown, we need to use the Student's t-distribution.
Calculate the standard error (SE):
SE = s / √n = 2.16 / √6 ≈ 0.88
Find the critical value (t*) for a 95% confidence level and (n-1) degrees of freedom. For a sample size of 6, the degrees of freedom (df) is 5. The critical value is approximately 2.571.
Calculate the margin of error (ME):
ME = t* * SE = 2.571 * 0.88 ≈ 2.26
Construct the confidence interval:
CI = x̄ ± ME
CI = 5 ± 2.26 = (2.74, 7.26)
Therefore, the 95% confidence interval for the mean of the population is (2.74, 7.26).
For questions 6, 7, and 8, the calculations are similar to those described above, but we use different formulas for the confidence interval for proportions instead of means. The formulas involve the sample proportion, margin of error, and critical values from the standard normal distribution.