9.5 Additional information and full hypothesis test examples

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• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset α .
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is α = 0.05.
• When you calculate the p -value and draw the picture, the p -value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis , ${H}_{a}$ , tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• H a never has a symbol that contains an equal sign.
• Thinking about the meaning of the p -value    : A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

The following examples illustrate a left-, right-, and two-tailed test.

H o : μ = 5, H a : μ <5

Test of a single population mean. H a tells you the test is left-tailed. The picture of the p -value is as follows:

Try it

H 0 : μ = 10, H a : μ <10

Assume the p -value is 0.0935. What type of test is this? Draw the picture of the p -value.

left-tailed test

H 0 : p ≤ 0.2   H a : p >0.2

This is a test of a single population proportion. H a tells you the test is right-tailed . The picture of the p -value is as follows:

Try it

H 0 : μ ≤ 1, H a : μ >1

Assume the p -value is 0.1243. What type of test is this? Draw the picture of the p -value.

right-tailed test

H 0 : p = 50   H a : p ≠ 50

This is a test of a single population mean. H a tells you the test is two-tailed . The picture of the p -value is as follows.

Try it

H 0 : p = 0.5, H a : p ≠ 0.5

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the p -value.

two-tailed test

Full hypothesis test examples

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

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.

Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Stats and press ENTER . Arrow down and enter 16.43 for μ 0 (null hypothesis), .8 for σ , 16 for the sample mean, and 15 for n . Arrow down to μ : (alternate hypothesis) and arrow over to< μ 0 . Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value ( p = 0.0187) but it also calculates the test statistic ( z -score) for the sample mean. μ <16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 ( p -value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off.

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 :

$P\left(\overline{x}<16\right)=$ 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.)

According to my understanding, statistics are numerical data or information presented in charts or graphs
thanks
Said
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thank you sir pragaph bejo
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