NHST for sample means of a continuous random variable with critical values

NHST with p-values

The previous section discussed how to conduct a null hypothesis significance test when you have a normally distributed continuous variable by computing a p-value from z-scores. However, a more computationally friendly version of this method is to compare the critical value (z-score from α) to the test statistic (z-score from data). This is a simple modification of the p-value method where you compare the z-score for your chosen alpha to the z-score for your data to inform you whether your p-value is less than or greater than your alpha. We will elaborate on this method here.

NHST with critical values

The steps, process, and interpretations for the critical value method of NHST is the same as the p-value method. Below we highlight the adjustments, in italics, you make:

Step 1: Determine H0 and Ha. Remember, they are contradictory, mutually exclusive and exhaustive.

This involves identifying the best distribution for your research: Is your outcome variable (your independent or criterion variable) a continuous normal variable? Is it a discrete binomial distribution, etc.?

This informs your hypotheses to know if you are stating hypotheses about continuous variable means or discrete distribution proportions or means, etc.

Step 2: State your decision criterion and identify if you have a left, right, or two-tailed test.

Alpha is stated to be .05 by default.

Based on your alternative hypothesis, determine if you have a left (<), right (>), or two-tailed test (≠).

Find and state your critical value. Given your stated alpha level and type of test (left, right, or two-tailed), determine your critical z-score value. For example, if you have an alpha of .05, you have one of the following options depending on your alternative hypothesis:

1) Right-tailed test: We find in the unit normal table a value of .05 in the tail corresponds to a z-score of 1.645. Thus, the critical z-score would be 1.645.

2) Left-tailed test: We find in the unit normal table a value of .05 in the tail corresponds to a z-score of -1.645. Thus, the critical z-score would be -1.645.

3) Two-tailed test: We find in the unit normal table a value of .025 in the tail corresponds to a z-score of 1.96. Thus, the critical z-scores would be 1.96 and -1.96 (one critical z-score for each tail).

Step 3: Collect, and identify, your data.

Step 4: Compute your test statistic.

Compute the z-score for your data (from step 3).

Notice this step is shorter and easier! We only need to compute the z-score for the data and do not need to find the associated p-value now.

Step 5: Make your conclusions.

Compare the observed z-score (computed in step 4) to the stated critical z-score (from step 2). If your observed z-score is more extreme than your stated critical z-score then we reject the null hypothesis and find support for the alternative. If the computed z-score is less extreme than our stated critical z-score, then we fail to reject the null hypothesis and do not find support for the alternative hypothesis. Interpret in terms of your stated hypotheses and variables.


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EXAMPLE 1

Revisiting the example of Jeffrey our freestyle swimmer, we will complete the NHST steps once more, using the critical value method rather than the p-value method.

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.

To begin, let's set up the Hypothesis Test:

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

Step 1: State the hypotheses

H0: μ ≥ 16.43  

There is no effect of the new googles on Jeffrey's swimming time.

Ha: μ < 16.43

There is an effect of the new googles on Jeffrey's swimming time: the new googles help Jeffry swim faster.

Step 2: State the decision criterion

Our decision criterion (alpha) will be .05.

The “<” tells you this is one-tailed test (it is a left-tailed test).

Given that we have a left-tailed test with an alpha of .05, we look in our unit normal table and find that the z-score that corresponds to the lower .05 of the distribution is -1.645. Thus, our critical z-score is 1.645.

Step 3: Collect the data

Jeffrey swan the 25-yard freestyle with the new googles 15 times and had a mean time of 16 seconds.

Step 4: Compute the test statistic

Compute the z-score for a mean of 16 with a sample of 15 swims:

Step 5: Make your conclusions and interpretations

Compare your computed z-score of -2.08 to your critical z-score of -1.645.

Since our computed z-value of -2.08 is more extreme (i.e., a larger negative value) than our critical z-score of -1.645, our statistic is in our critical region and we reject the null hypothesis. Alternately stated, the critical z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), we reject H0.

Thus, at the 5% significance level, we conclude that we find a significant effect of the googles such 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 graph below shows α, the p-value, and the test statistics and the critical value.

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Important to note

The conclusion of NHST with the p-value and critical value methods are the same, because it uses the exact same information and logic - the only difference is whether you are talking and comparing in terms or probabilities (which represent z-scores) or z-scores (which can be representing with probabilities)!

Lastly, and importantly, the critical z-score for a left-tailed test with an alpha of .05 will always be -1.645! Similarly, the critical z-score for a right-tailed test with alpha of .05 will always be 1.645, and for a two-tailed test with alpha of .05 it will be -1.96 and 1.96. This simplifies our process by making the unit normal table, and p-value computations in the NHST process - we just have to remember these two values (+/- 1.645 and +/- 1.96) when using an alpha of .05!


References

  1. https://courses.lumenlearning.com/introstats1/chapter/additional-information-and-full-hypothesis-test-examples/

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