9.5 Additional information and full hypothesis test examples  (Page 53/51)

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H ₀ : μ = 16.43   H _a : μ <16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: $\overline{X}$ = the mean time to swim the 25-yard freestyle.

Distribution for the test: $\overline{X}$ is normal (population standard deviation is known: σ = 0.8)

$\overline{X}~N\left(\mu ,\frac{{\sigma }_X}{\sqrt{n}}\right)$ Therefore, $\overline{X}~N\left(16.43,\frac{0.8}{\sqrt{15}}\right)$

μ = 16.43 comes from H ₀ and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( $\overline{x}$ <16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

Graph:

μ = 16.43 comes from H ₀ . Our assumption is μ = 16.43.

Interpretation of the p -value: If H ₀ is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > p -value, reject H ₀ .

This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.

Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The p -value can easily be calculated.

When the calculator does a Z -Test, the Z-Test function finds the p -value by doing a normal probability calculation using the central limit theorem :

$P(\overline{x}<16)=$ 2nd DISTR normcdf $\left(-10^99,16,16.43,0.8/\sqrt{15}\right)$ .

The Type I and Type II errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, onaverage, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)