A new medication that claims to improve typing ability is currently being tested. The average person types at 30 wpm(words per minute) with a standard deviation of 16. The medication is expected to increase average wpm to 46. A sample of 16 individuals is taken to determine if the medication improves typing ability. Use alpha = .05 to test this claim.
Let�s go ahead and assume that the medication works and really does increase wpm to 46. What is the probability of us correctly concluding this with our statistical test?
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Figure 1. |
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Power |
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Power is the probability of correctly rejecting the null hypothesis. |
Beta(�) |
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Beta(�) is the probability of incorrectly retaining the null hypothesis. |
For this example, let's first calculate the standard error of the mean so we can draw the distribution:
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Figure 2. |
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Now, let's draw the distribution:
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Figure 3. |
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If the null hypothesis is true, we expect to calculate a z that is somewhere between -1.96 and 1.96.
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Figure 4. |
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This new distribution(on the right) is what it would look like if the distribution really had a mean of 46. In this picture, the orange area refers to our power, while the blue area refers to our probability of making a Type II Error, assuming that the null hypothesis is false.
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Figure 5. |
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The z-score of 1.96 means that we are 1.96 standard errors above the mean. 1.96 refers to the value 37.84. We can now use 37.84 to calculate what the z-score would be for the graph on the right:
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Figure 6. |
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Looking up 2.04 in our z-table we find the area in the body to be 0.9793. This means:
Power = 0.9793
Type II Error = = 1 - 0.9793 = 0.0207
There is about a 98% of us correctly rejecting the null hypothesis, given that it is false. There is a 2% chance of us making a Type II Error.