Question

Assume that the 129 patients in the Patients dataset represent the entire population of interest. If you were interested in age of the patients and took a sample of 25 patients from this population, what is the standard error of the mean?   What if you took a sample 64 patients from this population, what is the standard error of the mean? What happens to the standard error of the mean as the sample size increases?  If you select a sample of 64 patients, what is the probability that the sample mean is below 75? How to calculate in excel.

Age (Years) 78 74 89 81 87 65 90 61 90 78 78 71 76 76 79 72 72 64 72 69 63 78 83 62 71 83 63 83 76 79 65 79 74 63 84 90 73 81 75 87 70 73 77 71 76 49 78 86 67 69 73 88 67 69 77 64 76 64 41 49 59 81 74 77 (Here is sample of 64 Patients)  

Answer 1

SE is standad error

we know tha the standard error of the mean is given as SE =

So, when the we take a sample of 25, then the Standard error will be . It become 1/5th of the standard deviation.

here is the standard deviation

if we take a sample of 64 patients, then the standard error will be SE = =

So, it become 1/8th of the standard deviation

It is clear that when we take a sample of 25, the standad error was 1/5th of standard deviation and when we take a sample of 64, the standad error was 1/8th of the standard deviation. So, as the sample size increases the standard error decreases.

Probability using the excel

Put the data in the excel sheet

Use formla "AVERAGE" to get the mean value, we get mean = 73.58

Use formula "stdev.s" to get the sample standard deviation, we get Sd = 9.99

Now divide the standard deviation by square root of 64 to get the standard error, we get SE = 9.99/sqrt(64) = 1.25

Use the following formula in next cell

t statistic =

place the given values of , we get

t statistic =

Now use the excel formula "TDIST(x, df, tails)"

put x = calculated t value = -1.14, df = degree of freedom = n-1 = 64-1 = 63 and tails = 1 for right tailed hypothesis

we get

p value = tdist(-1.14,63,1) = 0.1293

So, the required p value is 0.1293