Question 760399: A telecommunication company wants to estimate the mean length of time (in minutes) that 18-to-24 year olds spend text messaging each day. In a random sample of twenty-seven 18-to-24 year olds, the mean length of time spent text messaging was 29 minutes. From past studies, the company assumes that the population standard deviation is 4.5 minutes. Assume that text messaging times are normally distributed. Use the flowchart on Page 322 to decide whether to use the normal distribution or the t-distribution, and then construct the 90% confidence interval for the population mean s number of minutes 18-to-24 year olds spend text messaging each day.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size is equal to 27.
mean time is equal to 29 minutes.
population standard deviation is equal to 4.5 minutes.
text messaging times are assumed to be normal.
the population size is less than 30 but the population distribution is assumed to be normal so the z or the t-test would be appropriate.
since the population standard deviation is known, the z-test is the one that can be used.
with a sample siae of 27, the results should probably be comparable.
let's see what happens with the z-test.
the sample size is equal to 27
the population standard deviation is equal to 4.5 minutes.
the standard error is equal to 4.5 / sqrt(27) which is equal to .8660.
the 90% confidence interval would have a z-score from -1.6449 to 1.6449.
those are the critical z-values.
the critical x-values are the mean plus or minus the critical z-values times the standard error of the distribution of the mean.
they would become:
mean = 29
mean - (.8660 * 1.6449) = 27.5755
mean + (.8660 * 1.6449) = 30.4245
the confidence interval is between 27.5755 minutes and 30.4245 minutes.
now let's see what happens with the t-test.
you get the same standard error of .8660.
the critical t-values are based on the degrees of freedom which are equal to sample size minus 1 which makes the degrees of freedom equal to 26.
we look up the critical t-value for 90% confidence interval and 26 degrees of freedom to get:
critical t-value is equal to a t-score from -1.7056 to 1.7056
the critical x-values become:
mean = 29
mean - (.8660 * 1.7056) = 27.5230
mean + (.8660 * 1.7056) = 30.4770
the results are a little broader with the t-score than with the z-score.
as the degrees of freedom get higher, the t-score approaches the z-score.
theoretically, if the degrees of freedom are infinitely high, then the t-score and the z-score should converge.
for example, at the 90% confidence limit, the z-score is equal to plus or minus 1.6449.
at the 90% confidence limit, the t-score is equal to plus or minus the following at the following degrees of freedom.
degrees of freedom 90% confidence interval t-score
2 2.92
22 1.7171
222 1.6517
2222 1.6455
22222 1.6449
at 22222 degrees of freedom, the critical t-score is equal to the critical z-score.
you can see that, as the degrees of freedom go up, the t-score is approaching the z-score.
bottom line:
in this problem you would use the z-test because you have a normal distribution in the population and you know the population standard deviation.
if you did not know the population standard deviation, but had the sample standard deviation, you would have used the t-test.
the rules for when to use the z-test and when to use the t-test are muddy at best.
look for different sources of information and there is a high probability that you will get different answers.
one such source states:
1. Z-test is a statistical hypothesis test that follows a normal distribution while T-test follows a Student’s T-distribution.
2. A T-test is appropriate when you are handling small samples (n < 30) while a Z-test is appropriate when you are handling moderate to large samples (n > 30).
3. T-test is more adaptable than Z-test since Z-test will often require certain conditions to be reliable. Additionally, T-test has many methods that will suit any need.
4. T-tests are more commonly used than Z-tests.
5. Z-tests are preferred than T-tests when standard deviations are known.
other documents will tell you the following:
if n > 30 and population standard deviation is known, then use z-test.
if n < 30 and population assumed to have normal distribution and standard deviation is known, then use z-test.
the first document states that when n < 30, the t-test would normally be used.
bottom line - all very confusing.
since your population is assumed to be normal and your population standard deviation is known, i would go with the z-test.
as you can see, the results are not significantly different from each other.
this is because the sample size is close to 30 if not greater than 30.
that makes the z-score and the t-score approximate each other more closely than if the sample size were something very small, like 2 or 3, in which case the t-score and the z-score would be significantly different.
lot of wind here, but i would go with the z-test rather than the t-test on this one.
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