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Estimating a Population Mean for Large and Small Samples

Only for above OTA's. All others will be called in. Thank you.
I am having a lot of trouble understanding statistics even thoughI have studied. In an attempt to pull the pieces together,I am looking for the following to be explained. Please only accept this if you are willing to take the time to explain clearly all, a-f for each of the following.
a)Formulas b)when each would be used-what sort of problem to apply it to c)how to apply it, d) why it is applied to a particular problem, e)with math being explained on an elementary level f) real life sample.
Apply to each of the following:
Estimating a Population mean: Large samples
Estimating a population mean, Small samples
Determining sample size required to estimate u ( s.d.known and not known)
Estimating a population proportion
Estimating a population varience
Finding value of test statistic z
Testing a claim about a mean:Large samples
P-value method of testing hypothesis
Testing a claim about a proportion

Solution Preview

Please see attached file. Hope this helps.

1. Estimating a Population mean: Large samples:
Large-sample 100% confidence interval for a population mean,

where is the z-value that locates an area of to its right, is the standard deviation of the population from which the sample was selected, n is the sample size, and is the value of the sample mean.
Assumption: n ³ 30
[When the value of is unknown, the sample standard deviation s may be used to approximate in the formula for the confidence interval. The approximation is generally quite satisfactory when n ³ 30.]
Example: Suppose that in the previous year all graduates at a certain university reported the number of hours spent on their studies during a certain week; the average was 40 hours and the standard deviation was 10 hours. Suppose we want to investigate the problem whether students now are studying more than they used to. This year a random sample of n = 50 students is selected. Each student in the sample was interviewed about the number of hours spent on his/her study. This experiment produced the following statistics:
= 41.5 hours s = 9.2 hours
Estimate , the mean number of hours spent on study, using a 99% confidence interval. Interpret the interval in term of the problem.
Solution The general form of a large-sample 99% confidence interval for is

or (38.14, 44.86).
We can be 99% confident that the interval (38.14, 44.86) encloses the true mean weekly time spent on study this year. Since all the values in the interval fall above 38 hours and below 45 hours, we conclude that there is tendency that students now spend more than 6 hours and less than 7.5 hours per day on average (suppose that they don't study on Sunday).
2. Estimating a population mean, Small samples
Assumption required for estimating based on small samples (n < 30)
Small-sample confidence interval for

where the distribution of t based on (n - 1) degrees of freedom.
Example: Determine the t-value that would be used in constructing a 95% confidence interval for based on a sample of size n = 14.
Solution For confidence coefficient of .95, we have

We require the value of t.025 for a t-distribution based on (n - 1) = (14 - 1) = 13 degrees of freedom. In t-table, at intersection of the column labeled t.025 and the row corresponding to df = 13, we find the entry 2.160 (see Figure 7.6). Hence, a 95% confidence interval for , based on a sample of size n = 13 observations, would be given by

3. Determining sample size required to estimate u ( s.d. known and not known)
Choosing the sample size for estimating a population mean to within d units with probability

(Note: The population standard deviation will usually have to be approximated.)
Choosing the sample size for estimating a population proportion p to within d units with probability

where p is the value of the population proportion that we are attempting to estimate and q = 1 - p.
(Note: This technique requires previous ...

Solution Summary

The expert estimates a population mean for large and small samples. A population proportion is estimated.

$2.19