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December 21, 2014

December 21, 2014

Posted by **Cam** on Friday, May 3, 2013 at 11:45am.

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.

- Statistics -
**PsyDAG**, Friday, May 3, 2013 at 1:55pmWe 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).

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