The null hypothesis H0: mean=123 gallons
The alternative hypothesis Ha: mean≠123 gallons
Significance level ɑ=0.05 (2-tail), so we use ɑ/2 for each tail, because the mean could be more or less than 123 gallons.
Test statistic T=(119-123)/(5.3/√16)=-3.02.
This corresponds to a probability < 0.025.
The test statistic is less than (more extreme than) the critical p-value we conclude that there is sufficient evidence to reject H0, with a 95% degree of confidence, so it appears that the mean is no longer 123 gallons.
[In ordinary words, the almanac states the mean to be 123 gallons. We don’t have the population standard deviation which would tell us what sort of spread we could expect from this mean. That is, how much deviation we could expect by chance. But we do have a sample to work from. And this sample includes a standard deviation. We make an estimate of the population standard deviation by adjusting the sample standard deviation to make it narrower. Dividing by the square root of the sample size is the traditional way of doing this estimate.
And instead of using the normal distribution we make allowances for the relatively small size of the sample compared with the larger size of the population. This is the t-distribution which uses the degrees of freedom (in this case 15) to give us what looks like a normal distribution curve. For larger sample sizes the t-distribution looks more and more like the normal distribution.
Having established the tools we need to work with we can now find out if the sample mean of 119 is within the expected spread of data around a mean of 123.
This is where the significance or confidence level comes in. The distribution curve has little “tails” to the left and right. Although the area under the curve includes 100% of the data, we only need to be 95% sure (good enough!), that is, a significance of 5%. In this problem, we are only interested in whether the mean is 123 or not, so that includes smaller or greater than 123 gallons. We are limiting the spread or interval to a total width of 95%. Since the distribution curve is symmetrical, half of this spread (47.5%) will be to the left of the mean (lower than 123 gallons) and half will be to the right (higher than 123 gallons). The tails on either side each contain 2.5% of the data.
The sample has a mean of 119 gallons which is lower than 123. So we want to know whether 119 is within the expected range or spread of data where 123 is the mean. What we call the t test statistic is simple how many standard deviations from the mean of 123 gallons is the sample mean. It’s more than 3 standard deviations lower, and that corresponds to a probability less than 2.5% as measured by the left side of the distribution curve. That indicates that it’s extremely unlikely that by chance (within 95%) a sample would have a mean of 119 gallons. We are looking at the left, lower, 47.5%.
That leads us to conclude that 123 gallons is no longer the mean, which is the recorded almanac value and the basis of the null hypothesis. So the null hypothesis is rejected in favour of the alternative hypothesis that the mean has changed. We are 95% confident about that.]
