For this page we will assume that we have a large population and that we can take some measurement on that population. If you need some context for this we could say that the population is all current WCC male students and the measurement is the percent fat of the students as measured by a Beurer Glass Body Analysis Scale. [As a small note: getting "percent body fat" measurements with such a scale may or may not be all that accurate but that is not a concern here. What is a concern is that this is a measurement and it is a reproducable measurement. That is, if you measure your percent body fat, get off the scale and get back on, it will give you the same measurement.]

We have a null hypothesis that the mean percent body fat for current WCC male students is 24 and an alternative hypothesis that the mean percent body fat for current WCC male students is greater than 24. We could get all 6,000 plus current WCC male students to come in, get measured, and from that data we could determine the exact mean percent fat of that population. However this is not practical. Instead, we will take a random sample of 38 such students, take percent body fat measurements on those students, using the Beurer Glass Body Analysis Scale, and then look at the sample mean of those measurements. Based on that sample mean we will see if we have evidence to reject the null hypothesis in favor of the alternative hypothesis.

We will assume that the the standard deviation of percent fat of current male WCC students is unknown.

First Version:
For this version we will set our level of significance to be 5%. That is, as we judge the null hypothesis we are willing to be wrong in rejecting the null hypothesis 5% of the time.
So far we have:
`H_0: mu=24`
`H_1: mu>24`
`alpha=0.05`
`n=38`

Second Version:
The only thing we will change for the second version of the problem is our declared level of significance. For this version we say that we are only willing to incorrectly reject the null hypothesis in favor of the alternative hypothesis `1%` of the time. Everything else remains the same for this version so:
`H_0: mu=24`
`H_1: mu>24`
`sigma=3.5`
`alpha=0.01`
`n=38`

Critical Value Method: Because we have changed the level of significance, we need to compute a new t-value. On the TI-84 we could use the command invT(0.99,37) to get the t-value 2.4314. On any of the calculators we could use the ASTUDT program to get the t-value 2.4314. Having found that value, we take our sample of 38 students and get the sample standard deviation of 3.71. Therefore, we can set our critical value at `24+2.4314(3.71/sqrt(38))=25.4633`. then, we compare the sample mean of 25.032 to that critical value. In this case, the sample mean is not more extreme than the critical value. Therefore, we do not reject the null hypothesis; in effect we accept the null nupothesis.

Attained Significance Level Method: Just as above, we take our sample and compute the sample statistics finding `barx=25.032` and `S_x=3.71`. Then we can ask "How strange is it to get a sample mean that is this extreme or even more extreme for a distribution that is a Student's-T with 37 degrees of freedom. As before, we have the long and short way to do this. We can use our calculator to convert our sample mean to a t-value from a population that has mean=0 and standard deviation=1. We do this by evaluating `(bar_x-mu_0)/(S_x/sqrt(38))=25.032-24)/(3.71/sqrt(38))=1.7148`, precisely the same calculation we did in version one. Then we use the tcdf( command, tcdf(1.7148,999,37) to find the probability of getting such a value, or a more extreme value, for our given sample size. As before this is `4.737%`. Because this version is being considered at the `1%` level of significance, we do not have evidence to reject the null hypothesis, which means that we accept the null hypothesis. The more straight forward solution to this is to use the T-Test... command. This will be exactly the same process that we did with version one. The input and output screens are shown below: That resulting screen shows the same results that we got via the longer method. Therefore, the conclusion is the same.

Third Version:

For the third version of this problem we will return to our original problem but change the sample size from 38 to 87. Thus, the significant values are:
`H_0: mu=24`
`H_1: mu>24`
`alpha=0.05`
`n=87`

Critical Value Method: The change from a sample of size 38 to one of size 87 means that we need to get a new value for the standard deviation of the sample mean. We draw our random sample, compute the sample standard deviation, which we find to be 3.71, and then use that to compute the standard deviation of the sample mean, namely, `3.7/sqrt(87)=0.3978`. Returning to the first version we know that the t-value, for a population with mean=0 and standard deviation of 1, that has 5% (our level of significance) of the area to the right of that value is 1.6871. We translate that to our particular sample by computing `1.6971*S_barx+mu_0=(1.6871)(0.3978)+24=24.673` and that becomes our critical value. From our sample we compare the sample mean and find that `barx=25.032`. We compare that to the critical value We see that our sample mean, `barx=25.032`, is more extreme than the critical value. Therefore, based on our sample, we reject the null hypothesis in favor of the alternative hypothesis.

Attained Significance Level Method: Having seen, in the earlier versions, the long way to do this, we will just stick with the short way. We take a random sample, find the sample mean and the sample standard deviation: `barx=25.032` and `S_x=3.71`. Then we use the T-Test... command. The input and output screens follow: With this new sample size we know that This tells us that for samples of our size from a population, if the null hypothesis, `H_0: mu=24`, is true, then we would expect to get a sample mean of `25.032` or higher in only `0.557%` of the such samples. Because this is less than our stated level of significance, namely, `alpha=0.05`, we reject the null hypothesis in favor of the alternative hypothesis.

Fourth Version:

For the fourth version of this problem we will change the third version to use a level of significance set to `1%`. As in the third version, we will still have a samle size of 87. . Thus, the significant values are:
`H_0: mu=24`
`H_1: mu>24`
`alpha=0.01`
`n=87`

Critical Value Method: Again, we draw our sample and find the `xbarx=25.032` and `S_x=3.72`. From that we compute `S_barx=3.71/sqrt(87)=0.39781`. The t-value that has `99%` of the area to its left, is 2.4314 (just as in version two). Thus, our critical value becomes `24+(2.4314)(0.39781)=24.967`. From our sample we had the sample mean, `barx`, is `25.032`. This value is more extreme than our critical value. That is to say, because we are only willing to be wrong `1%` of the time, and because we know that `99%` of all samples of our size taken from the population will have a sample mean that is less than or equal to `24.967`, finding a sample that has `barx=25.032` falls outside of that group of `99%` of all samples. It is just too extreme. If the null hypothesis were true, we would get such a random sample less than `1%` of the time. Therefore, based on our sample, we reject the null hypothesis in favor of the alternative hypothesis.

Attained Significance Level Method: We take our sample and we find that the sample mean is `25.032` and the sample standard deviation is 3.71. We can use our calculator to answer our question by using the T-Test... command. This will have the same input and output screens as in version three on this page. This tells us that for samples of our size, if the null hypothesis, `H_0: mu=24`, is true, then we would expect to get a sample mean of `25.032` or higher in only `0.557%` of the such samples. Because this is less than our stated level of significance, namely, `alpha=0.01`, we reject the null hypothesis in favor of the alternative hypothesis.

Fifth Version:

For the fifth version of this problem we will return to our original problem but change the sample size from 38 to 14. Note that having a smaple size less than 30 brings into question the extent to which the population of sample means (from repeated samoling of size 14) conformas to a normal distribution. If we have reason to believe that the underlying poputation is normally distributed, then we can use such a small sample size. We will continue using the assumption that the underlying population really does have a normal distribution of the percent fat of current male WCC students. Thus, the significant values are:
`H_0: mu=24`
`H_1: mu>24`
`alpha=0.05`
`n=14`

Critical Value Method: We take our sample and find `barx=25.032` and `S_x=3.71`. The t value that has `95%` of the area to its left is, as in version one above, 1.6871. The change in sample size means that we now compute `S_barx=3.71/sqrt(14)=.9915`. Thus, our translated critical value will be `24+(1.6871)(.9915)=25.673`. we compare our sample mean `25.032` to this critical value and find that the sample mean is not nore extreme than our critical value. Therefore, based on our sample, we can not reject the null hypothesis in favor of the alternative hypothesis. In essence, we are left with accepting the null hypothesis.

Attained Significance Level Method: We draw out sample, getting `barx=25.032` and `S_x=3.71` and place those values into the T-Test... command yielding: This shows that the likelihood of getting a result as extreme or more extreme than our 25.032 is 15.85%. Thus, if the null hypothesis is true, we should not be "shocked" to see a sample such as ours. Therefore, we do not have evidence that is weird enough for us to reject the null hypothesis. We abbreviate this by saying that we accept the null hypothesis.