1. Calculate the margin of error if the necessary requirements are satisfied. Please assume the original population is normally distributed: a. Confidence Level = 95%, n = 78, σ = 6; E = ?

2. A concert promoter wants to estimate the mean age of concertgoers who live in the Chicago area. In a random sample of 70 concertgoers, the mean age is found to be 27.8 years. From past studies, the standard deviation is known to be 1.6 years. Construct a 99% interval for the population

3. You want to conduct a study to determine if people would like to purchase dinner vouchers with their tickets to a particular sporting event. You decide to use a survey. How many people should you ask to complete the survey if you want to be 90% confident that the sample mean is within 3 points from the population mean? From previous studies, the standard deviation is 95.

4. Construct the indicated confidence interval for the population mean μ using a t-distribution: C = 99%, m = 46.9, s = 7, n =9

5. A survey of 1500 adults found that 445 of them own a pet cat. Construct a 95% confidence interval of the true proportion of adults who own a cat.

6. A researcher wants to estimate, with a 95% confidence level the proportion of people who own 3 or more cars. A previous study showed that 23% of people asked owned 3 or more cars. The researcher wants to be accurate within 4%. Find the number of people he needs to ask to meet these criteria.

1. To calculate the margin of error we need to find the critical value corresponding to the required confidence level of 95%. We want the z value that's halfway between 95 and 100%, which is 97.5%. When we look this up under normal distribution we get 1.96, which means that the margin is 1.96 standard deviations from the mean. But we also have to consider the sample size of 78. There are 77 degrees of freedom (dof) for this sample size and this adjusts the z value, which applies to a very large sample, and we use the t distribution table, which gives us a value of 1.993. And we must account for the sample size in the standard deviation making it 6/√77=0.684 approx. Now we apply the t value=0.684*1.993=1.36 approx. So E=±1.36.
2. 99% corresponds to a t value of 99.5% with 69 dof =2.90. Margin of error=2.90*1.60/√70=0.56 years. Applying this value to the mean we get 27.8±0.56=interval of 27.2 to 28.4 years for the population. This means we can be 99% sure that the average age is between these limits.
4.
answered Oct 20, 2016 by Top Rated User (415,140 points)

3.

90% corresponds to a t value with unknown dof=1.645 which applies to a large population. E=±3 and SD=95. We know the sample SD=s/√n where n is the sample size. E=±st√n so 3=95*1.645/√n. Therefore n=(95*1.645/3)^2=2714 approx. The dof for this sample size can be considered large enough for t=1.645 to be valid.

4. C=99% for dof=8 gives t=3.355. E=7*3.355/√9=7.83 so the pop mean=46.9±7.83 which is the interval 39.1 to 54.7.

5. The t value for 95% with 1499 dof is 1.96 approx. The mean of the sample is 445/1500=0.297 approx (297 in 1000). We don't know the SD, so we use the normal distribution table and the z value for 97.5%=1.96 (same as the t value for 95%). This tells us that mean/SD=1.96 so the estimated SD=0.297/1.96=0.15 approx. The sample size is 1500 so estimated pop SD=0.15/√1500=0.004 approx. E=±1.96*0.004=0.008 approx. Therefore the mean is 0.297±0.008 or 297±8 in every 1000 adults. This comes to 445±12 in 1500 adults.

6.

For this part of the question we have t and z values of 1.96 corresponding to the 95% required confidence level. As in part 3 we have to find n. The mean is 23%=0.23 and the estimated SD is 0.23/1.96=0.12 approx. Estimated pop SD=0.12/√n so E=±0.12t/√n and we want this to have a value of 4%=0.04, so 0.04=0.12t/√n and n=(0.12t/0.04)^2=9t^2. So the researcher needs to ask at least 9t^2 people.

If t is approximately 2 then 9t^2=36. As it happens t=2.03 with dof=36, so 37 people should be surveyed. Let's see if this fits the requirements.

E=±2.03*0.12/√37=0.04=4% so the results are expected to be 23±4%.