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#6 – Confidence Interval of Means – Activity 2
Name_____________________
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 (S=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?
2. A university wants to know more about the knowledge of students regarding international events. The
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, S=5). A sample of
30 students are tested (sample mean=58, Standard Error=3.2). Compute a 99 percent confidence interval
based on this sample's data. How do these students compare to the national sample?
3. 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
7.5
7 14.5
9 10.5
9
9 3.5 8 11 7.5 18 20
15 19 2.5
5 9 8.5 14 20
8
Compute a 95 percent confidence interval. What does this tell us?
4. 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.
a. 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?
b. 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?
c. If we find that that 0.2 cm is the standard deviation of the tail lengths of the newts in our sample
the population, then what is a 90% confidence interval for the population mean?
d. If we find that that 0.2 cm is the standard deviation of the tail lengths of the newts in our sample
the population, then what is a 95% confidence interval for the population mean?
Answers
1. The standard error of the mean is 1.807392228.
t=2.000
Confidence Interval at 95 percent: 44.4 < population mean < 51.6
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 44.4 minutes and 51.6 minutes.
There is a 5 percent chance than the population mean for stage IV sleep in Alzheimer's patients is less
than 44.4 minutes or more than 51.6 minutes.
2. t=2.756
Confidence Interval at the 99 percent: 49.2 < population mean < 66.8
While the data for these students are low in relation to the national scores, the constucted interval does
include the national mean. Therefore, the university may be on par with other universities. The
university may with to replicate their survey to further validate their results.
3. Mean = 10.92
Standard Deviation = 5.598265777
Standard Error = 1.167319108
t = 2.069
Confidence Interval at 95 percent: 8.50 < population mean < 13.33
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 8.5 hours and 13.33 hours. There is a 5
percent chance that the population mean does not lie within that interval.
4. 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.
C)
Here we do not know the population standard deviation, only the sample standard deviation.
Thus we will use a table of t-scores. When we use a table of t scores we need to know how many
degrees of freedom we have. In this case there are 24 degrees of freedom, which is one less than
sample size of 25. The value of t that corresponds to a 90% confidence interval is 1.71. By using
the formula for the margin of error we have a confidence interval of 5 – 1.71(0.2/5) to 5 +
1.71(0.2/5). After carrying out the arithmetic we have 4.932 cm to 5.068 cm as a confidence
interval for the population mean.
D) Here we do not know the population standard deviation, only the sample standard deviation.
Thus we will again use a table of t-scores. There are 24 degrees of freedom, which is one less
than sample size of 25. The value of t that corresponds to a 95% confidence interval is 2.06. By
using the formula for the margin of error we have a confidence interval of 5 – 2.06(0.2/5) to 5 +
2.06(0.2/5). After carrying out the arithmetic we have 4.912 cm to 5.082 cm as a confidence
interval for the population mean.