6.114 Effect of changing the alternative on power. The Statistical Power applet

Question

6.114 Effect of changing the alternative on power. The Statistical Power applet illustrates a power calculation similar to that in Figure 6.16 (page 404). Open the applet and keep the default settings for the null ( = 0) and the alternative ( > 0) hypotheses, the sample size (n = 10), the standard deviation ( 1), and the significance level ( = 0.05). In the "alt -box enter the value 1. What is the power? Repeat for alter- native equal to 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. Make a table giving and the power. What do you conclude? 6.115 Other changes and the effect on power. Refer to the previous exercise. For each of the following

Explanation / Answer

6.115

(a) If we change it to two sided alternative then

We will committ type II error if we failed to reject the null hypothesis even if it is false.

we will failed to reject the null hypothesis if x < H + Z97.5% (/n)

or if x < 0 + 1.96 * 1/10 => x < 0.62

so Pr(Type II error) = Pr(x < 0.62 ; 1 ; 1/10) = Pr(x < 0.62 ; 1 ; 0.3162)

Z = (0.62 - 1)/0.3162 = -1.20

Power = 1 - Pr(Type II error) = 1 - Pr(Z < -1.20) = 1 - 0.1151 = 0.8849

so, we see that power is reduced here if we choose two sided alternative

Now for any given true

Power = 1 - Pr(Type II error) = 1 - NORM (x < 0.62 ; ; 0.3162)

Making the table here

(b) Here increase   to 2

it will increase standard error of mean

we will failed to reject the null hypothesis if x < H + Z95% (/n)

or if x < 0 + 1.645 * 2/10 => x < 1.04

so Pr(Type II error) = Pr(x < 1.04 ; 1 ; 2/10) = Pr(x < 0.62 ; 1 ; 0.6325)

Z = (1.04 - 1)/0.6325 = 0.06

Power = 1 - Pr(Type II error) = 1 - Pr(Z < 0.06) = 1 - 0.5254 = 0.4746

so, we see that power is reduced here if we increase sigma to 2.

Now for any given true

Power = 1 - Pr(Type II error) = 1 - NORM (x < 1.04 ; ; 0.6325)

(c) If we increase n from 10 to 20, it will reduce standard error and that will increase the power.

we will failed to reject the null hypothesis if x < H + Z95% (/n)

or if x < 0 + 1.645 * 1/20 => x < 0.3678

so Pr(Type II error) = Pr(x < 0.3678 ; 1 ; 1/20) = Pr(x < 0.3678 ; 1 ; 0.2236)

Z = (0.3678 - 1)/0.2236 = -2.83

Power = 1 - Pr(Type II error) = 1 - Pr(Z < -2.83) = 1 - 0.0023 = 0.9977

so, we see that power isincreased if we increase n from 10 to 20

Now for any given true

Power = 1 - Pr(Type II error) = 1 - NORM (x < 0.3678 ; ; 0.2236)

m Power 0.1 0.0501 0.2 0.0922 0.3 0.1559 0.4 0.2435 0.5 0.3524 0.6 0.4750 0.7 0.6001 0.8 0.7156 0.9 0.8122 1 0.8854

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