Confidence Interval of Means – Activity 2Name______

  1. A sample of Alzheimer's patients are tested to assess the amount of time in stage IV sleep. It has been hypothesized that individuals sufferering from Alzheimer's Disease may spend less time per night in the deeper stages of sleep. Number of minutes spent is Stage IV sleep is recorded for sixty-one patients. The sample produced a mean of 48 minutes (assume σ=14 minutes) of stage IV sleep over a 24 hour period of time. Compute a 95 percent confidence interval for this data. What does this information tell you about a particular individual's (an Alzheimer's patient) stage IV sleep?
  1. A university wants to know more about the knowledge of students regarding international events. They are concerned that their students are uninformed in regards to new from other countries. A standardized test is used to assess students knowledge of world events (national reported mean=65, σ=5). A sample of 30 students are tested (sample mean=58). Compute a 99 percent confidence interval based on this sample's data. How do these students compare to the national sample?
  1. A sample of students from an introductory psychology class were polled regarding the number of hours they spent studying for the last exam. All students anonymously submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one section of the course polled. The data was used to make inferences regarding the other students taking the course. There data are below:

4.5 / 22 / 7 / 14.5 / 9 / 9 / 3.5 / 8 / 11 / 7.5 / 18 / 20
7.5 / 9 / 10.5 / 15 / 19 / 2.5 / 5 / 9 / 8.5 / 14 / 20 / 8

Assume σ=4.8. Compute a 95 percent confidence interval. What does this tell us?

  1. We start with a simple random sample of 25 a particular species of newts and measure their tails. The mean tail length of our sample is 5 cm.
  2. If we know that 0.2 cm is the standard deviation of the tail lengths of all newts in the population, then what is a 90% confidence interval for the population mean?
  3. If we know that 0.2 cm is the standard deviation of the tail lengths of all newts in the population, then what is a 95% confidence interval for the population mean?

Answers

  1. The standard error of the mean is 1.793.

Z*=1.96

Confidence Interval at 95 percent: 44.49population mean < 51.51

We are 95 percent sure that the population mean for the number of hours an Alzheimer's patient will spend in stage IV sleep in a 24 period of time is somewhere between 44.49 minutes and 51.51 minutes. (There is a 5 percent chance than the population mean for stage IV sleep in Alzheimer's patients is less than 44.49 minutes or more than 51.51 minutes.)

  1. Z*=2.576 SE=0.913

Confidence Interval at the 99 percent: 55.65 < population mean < 60.35

The data for these students are low in relation to the national scores, and the constructed interval does not include the national mean. Therefore, the university may be low performing in comparison with other universities.

  1. Mean = 10.92
    Standard Error = 0.9798

Z* = 1.96

Confidence Interval at 95 percent: 8.996 < population mean < 12.837

We are 95 percent sure that the actual population mean for the number of hours introductory psychology students studied for the last exam was somewhere between 8.996 hours and 12.837 hours. There is a 5 percent chance that the population mean does not lie within that interval.

  1. A)Since we know the population standard deviation, we will use a table of z-scores. The value of z that corresponds to a 90% confidence interval is 1.645. By using the formula for the margin of error we have a confidence interval of 5 – 1.645(0.2/5) to 5 + 1.645(0.2/5). (The 5 in the denominator here is because we have taken the square root of 25). After carrying out the arithmetic we have 4.934 cm to 5.066 cm as a confidence interval for the population mean.

B)Since we know the population standard deviation, we will use a table of z-scores. The value of z that corresponds to a 95% confidence interval is 1.96. By using the formula for the margin of error we have a confidence interval of 5 – 1.96(0.2/5) to 5 + 1.96(0.2/5). After carrying out the arithmetic we have 4.922 cm to 5.078 cm as a confidence interval for the population mean.