SOLUTION: Nielsen Media Research wants to estimate the mean amount of time (in minutes) that full-time college students spend watching television each weekday. Find the sample size necessary

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Question 761574: Nielsen Media Research wants to estimate the mean amount of time (in minutes) that full-time college students spend watching television each weekday. Find the sample size necessary to estimate that mean with a 15 minute margin of error. Assume that a 99% confidence level is desired. Also assume that a pilot study showed that the standard deviation of the population is 112.2 minutes.Thank you!
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
standard deviation of the population is equal to 112.2
confidence level is equal to .99 which makes alpha equal to .01
since you are dealing with a 2 tailed distribution, divide alpha by 2 and you get alpha / 2 = .005.
find the critical z value for an alpha of .005.
it is equal to 2.5758.
the following picture shows you that.
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the graph is showing you that the critical z-value is +/- 2.5758.
.005 of the area under the distribution curve is to the left of -2.5758.
.005 of the area under the distribution curve is to the right of 2.5758.
the sum of those 2 areas is equal to .01 which is your desired alpha.
the formula for margin of error is moe = czv * se
moe = margin of error
czv = critical z-value
se = standard error.
since you know moe and you know czv, you can solve for se.
the formula of moe = czv * se becomes 15 = 2.5758 * se
solve for se to get se = 15 / 2.5758 = 5.8234
the formula for standard error is se = sdp / sqrt(n)
se = standard error
sdp = standard deviation of the population.
n = sample size.
since you know sdp and se, the formula of se = sdp / sqrt(n) becomes 5.8234 = 112.2 / sqrt(n)
solve for sqrt(n) to get sqrt(n) = 112.2 / 5.8234 = 19.2671
square both sides of this equation to get n = 371.2209
that's your solution.
you now have the following information:
n = 371.2209
sdp = 112.2
se = 5.8234
czv = +/- 2.5758
the following picture shows you the area under the distribution curve when you use the critical z-value of 2.5758.
$$$$$$
you can find the confidence interval from this information if you know the mean.
assuming the mean is equal to 25, then you can calculate the confidence interval to be 25 +/- the margin of error.
that makes the confidence interval from 10 to 40.
the following picture shos you that the area outside the interval from 10 to 40 is equal to .01 if you assume the mean is 25 and you assume the standard error is equal to 5.8234.
the graph shows it as standard deviation, but it's really standard error because we are dealing with a distribution of sample means.
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