Question

A species of marine arthropod lives in seawater that contains calcium in a concentration of 32 mmole/kg of sea water. Thirteen of the animals are collected and the calcium concentration in coelomic fluid are determined. Results are summarized in the table below. A researcher plans to use these data to test H0: (mu) = 32 versus HA: (mu) (DNE) 32 at a significance level of 0.05, where (mu) = the mean calcium concentration in this arthropod’s coelomic fluid.

Mean	29.76923
Median	30
Standard Deviation	1.786703
Sample Variance	3.192308
Range	6
Minimum	27
Maximum	33
Sum	387
Count	13

4. Compute the power of the test when (mu) = 31.5.
5. Determine the sample size necessary in order to achieve a power of 80% when (mu) = 31.5.

Answer 1

4.

We need to compute the probability of a type II error (β), as well as the statistical power, for this two-tailed test, for the given significance level of α=0.05, and a sample size of n = 13. The critical value in this case is z_c = 1.96

In this case, the effect size passed is The statistical power will be computed using this value of the effect size d = 0.96869, α=0.05, n = 13

For all the purposes, we will assume that we are testing for one population mean, and we are using the z-statistics. This is a simplification that makes the calculations simpler.

First we compute the probability of making a type II error (β). β is the probability of rejecting a false null hypothesis. We will use the following notation: μ0​ is the hypothesized value of the population mean, and μ is the true population value.

The probability of making a type II error (\betaβ) is computed as follows:

But, if the true population mean is μ, then ​ has a standard normal distribution. Therefore, the following is obtained:

Observe that we have assumed without loss of generality that by symmetry, because if it is not the case, we would use the other tail to get the exact same result.

Now, we plug the values we know to get:

Now that we know β, we get the power directly as:

5.

From the previous equation can be observed that the value of β depends on n. Indeed, the value of β will decrease as n increases. In order to find the minimum n to reach the target power, we need to construct the following table:

Sample Size (n)	Lower (a = -zc-dѴn)	Upper (b = zc-dѴn)	β = P(a ≤ Z ≤ b)	Power = 1 - β
1	-2.9287	0.9913	0.8375	0.1625
2	-3.3299	0.5901	0.722	0.278
3	-3.6378	0.2822	0.611	0.389
4	-3.8974	0.0226	0.509	0.491
5	-4.1261	-0.2061	0.4184	0.5816
6	-4.3328	-0.4128	0.3399	0.6601
7	-4.5229	-0.6029	0.2733	0.7267
8	-4.6999	-0.7799	0.2177	0.7823
9	-4.8661	-0.9461	0.1721	0.8279
9	-4.8661	-0.9461	0.1721	0.8279
10	-5.0233	-1.1033	0.135	0.865
11	-5.1728	-1.2528	0.1051	0.8949
12	-5.3156	-1.3956	0.0814	0.9186

Therefore, looking at the table above, we find the minimum value of n that exceeds the minimum required target statistical power is n = 9.

=Pr(−zc​−dn​≤Z≤zc​−dn​)