If we take k samples from a normal distribution with fixed unknown mean and variance, and if we compute the sample mean and sample variance for these k samples, then the t-distribution (for k) can be defined as the distribution of the location of the true mean. In this way the t-distribution can be used to estimate how likely it is that the true mean lies in any given range.

If the data are normally distributed, the one-sided (1 − a)-upper confidence limit (UCL) of the mean, can be calculated using the following equation:

$\mathrm{UCL}_{1-a} = \overline{X}_n + t_{a,n-1}\frac{S_n}{\sqrt{n}}.$

In the given problem, we have the following values:

· Number of samples taken=20.

· Sampling mean=375

· Sample variance =130

We have been asked to test the following hypothesis: “the mean distance is a least 425 meters”. We are to use the hypothesis procedure to determine if there is sufficient evidence at a level of significance of 0.025 to refute the given hypothesis.

First let’s calculate the standard deviation of the sampling mean, which equals the following: s/(square root of n) = 130/(4.47) = 29.07

Since the sample size is small, we will use a t-distribution table instead of a standard normal distribution table. The value of t is equal to (value – sampling mean)/(standard deviation of the sampling mean).

Now we can go back to our UCL formula and solve the problem:

$\mathrm{UCL}_{1-a} = \overline{X}_n + t_{a,n-1}\frac{S_n}{\sqrt{n}}.$

UCL_1-a = 375 + 2.093*29.07           (Note: you can find t_{0.025, 19} = 2.093 in a t-distribution table for “one tail”).

UCL_1-a = 435.84

The upper confidence limit of the mean distance is 435.86, which means there is only a 2.5% chance that the mean distance is above 435.86. That means that there is a slightly greater than 2.5% chance that the mean distance is above 425. Therefore, we do not have sufficient evidence at level of significance of 0.025 that the mean distance is above 425.