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.

you didnt solve #8 mad dude

1. 99% = mean ± 2.575 SEm

SEm = SD/√n

2. ± 1.96 SEm

3. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.05) to get the value for the Z score.

4. Same table, but look for .10.

To find the confidence interval for the mean of a population, you can use the formula:

Confidence Interval = mean ± (critical value) * (standard deviation / √n)

Where:
- mean is the sample mean
- critical value represents the number of standard deviations that correspond to a specific confidence level (e.g., 99%, 95%, or 90%)
- standard deviation is the population standard deviation (or the sample standard deviation if it's not provided)
- n is the sample size

Now, let's calculate the confidence intervals for each of the given scenarios:

1. Given: n = 40, mean = 15, standard deviation = 0.6, and we need a 99% confidence interval.

To calculate the critical value, you need to determine the z-score associated with the desired confidence level. For a 99% confidence level, the critical value is approximately 2.576.

Using the formula:
Confidence Interval = 15 ± (2.576) * (0.6 / √40)
Confidence Interval = 15 ± (2.576) * (0.0949)
Confidence Interval ≈ 15 ± 0.2445

Therefore, the 99% confidence interval for the mean of the population is approximately [14.755, 15.245].

2. Given: n = 50, mean = 6, standard deviation = 1, and we need a 95% confidence interval.

For a 95% confidence level, the critical value is approximately 1.96.

Using the formula:
Confidence Interval = 6 ± (1.96) * (1 / √50)
Confidence Interval = 6 ± (1.96) * (0.1414)
Confidence Interval ≈ 6 ± 0.2773

Therefore, the 95% confidence interval for the mean of all pizza delivery workers is approximately [5.7227, 6.2773].

3. Given: n is not provided, mean = 2.3, standard deviation = 0.4, and we need a 90% confidence interval.

Since the sample size (n) is not given, you cannot calculate the exact confidence interval without it.

4. Given: n = 100, mean = 1.5, standard deviation = 0.3, and we need an 80% confidence interval.

For an 80% confidence level, the critical value is approximately 1.282.

Using the formula:
Confidence Interval = 1.5 ± (1.282) * (0.3 / √100)
Confidence Interval = 1.5 ± (1.282) * (0.03)
Confidence Interval ≈ 1.5 ± 0.0385

Therefore, the 80% confidence interval for the mean of the population is approximately [1.4615, 1.5385].

Please note that without knowing the sample size (n) in question 3, the confidence interval cannot be determined accurately.

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%

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.