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.
1. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005) and its Z score.
99% = mean ± Z SEm
SEm = SD/√n
2. Same but use .025.
3. Same but use .05. What is size of sample?
4. Same but use .10.
To find the confidence intervals for the mean of a population, we can use the formula:
Confidence Interval = Sample Mean ± (Critical Value * Standard Error)
The critical value corresponds to the confidence level and the degrees of freedom, while the standard error is calculated from the sample standard deviation and the sample size.
Let's go through each question and find the confidence intervals step by step:
1. For a 99% confidence interval with a sample of 40 people, the critical value can be found using a t-distribution table or a statistical software. Since the sample size is relatively large, a z-score can also be used. The critical value for a 99% confidence level (two-tailed test) is approximately 2.576.
The standard error is calculated as the sample standard deviation divided by the square root of the sample size. In this case, the standard error is 0.6 / sqrt(40) ≈ 0.0949.
So, the 99% confidence interval is:
15 ± (2.576 * 0.0949) = (14.757, 15.243)
2. For a 95% confidence interval with a sample of 50 pizza delivery workers, we'll use a z-score since the sample size is relatively large. The critical value for a 95% confidence level (two-tailed test) is approximately 1.96.
The standard error is 1 / sqrt(50) = 0.1414.
Therefore, the 95% confidence interval is:
6 ± (1.96 * 0.1414) = (5.733, 6.267)
3. For a 90% confidence interval with a given average number of TVs in each household (sample mean) of 2.3, we can calculate the critical value using a z-score table. For a 90% confidence level (two-tailed test), the critical value is approximately 1.645.
The standard error is 0.4 / sqrt(n), where n is the sample size. Note that the sample size is not given in the question, which makes it impossible to provide a specific confidence interval.
If the sample size is known, substitute it into the standard error formula and calculate the interval as:
2.3 ± (1.645 * standard error)
4. For an 80% confidence interval with a sample mean of 1.5, we need to find the critical value. Using a t-distribution table or statistical software, the critical value for an 80% confidence level (two-tailed test) with 100 degrees of freedom is approximately 1.984.
The standard error is 0.3 / sqrt(100) = 0.03.
Thus, the 80% confidence interval is:
1.5 ± (1.984 * 0.03) = (1.443, 1.557)
Remember, the confidence intervals provide a range within which we are confident that the population mean falls. These intervals indicate the level of precision and certainty in our estimates based on the sample data.