A company runs a manufacturing process that fills empty bottles with 100 mL of liquid. To monitor quality, the company randomly selects several bottles and measures the volume of liquid inside.

Determine the sample size the company must use for a t-test to detect a difference between 100 mL and 102 mL with a power of 0.80. Assume that a standard deviation is 5 mL.

nout = sampsizepwr('t',[100 5],102,0.80)

nout = 52

The company must test 52 bottles to detect the difference between a mean volume of 100 mL and 102 mL with a power of 0.80.

Generate a power curve to visualize how the sample size affects the power of the test.

nn = 1:100;
pwrout = sampsizepwr('t',[100 5],102,[],nn);

figure;
plot(nn,pwrout,'b-',nout,0.8,'ro')
title('Power versus Sample Size')
xlabel('Sample Size')
ylabel('Power')

An employee wants to buy a house near her office. She decides to eliminate from consideration any house that has a mean morning commute time greater than 20 minutes. The null hypothesis for this right-sided test is H0: $$\mu$$ = 20, and the alternative hypothesis is HA: $$\mu$$ > 20. The selected significance level is 0.05.

To determine the mean commute time, the employee takes a test drive from the house to her office during rush hour every morning for one week, so her total sample size is 5. She assumes that the standard deviation, $$\sigma$$, is equal to 5.

The employee decides that a true mean commute time of 25 minutes is too different from her targeted 20-minute limit, so she wants to detect a significant departure if the true mean is 25 minutes. Find the probability of incorrectly concluding that the mean commute time is no greater than 20 minutes.

Compute the power of the test, and then subtract the power from 1 to obtain $$\beta$$.

power = sampsizepwr('t',[20 5],25,[],5,'Tail','right');
beta = 1 - power

beta = 0.4203

The $$\beta$$ value indicates a probability of 0.4203 that the employee concludes incorrectly that the morning commute is not greater than 20 minutes.

The employee decides that this risk is too high, and she wants no more than a 0.01 probability of reaching an incorrect conclusion. Calculate the number of test drives the employee must take to obtain a power of 0.99.

nout = sampsizepwr('t',[20 5],25,0.99,[],'Tail','right')

nout = 18

The results indicate that she must take 18 test drives from a candidate house to achieve this power level.

The employee decides that she only has time to take 10 test drives. She also accepts a 0.05 probability of making an incorrect conclusion. Calculate the smallest true parameter value that produces a detectable difference in mean commute time.

p1out = sampsizepwr('t',[20 5],[],0.95,10,'Tail','right')

p1out = 25.6532

Given the employee's target power level and sample size, her test detects a significant difference from a mean commute time of at least 25.6532 minutes.

Compute the sample size, n, required to distinguish p = 0.30 from p = 0.36, using a binomial test with a power of 0.8.

napprox = sampsizepwr('p',0.30,0.36,0.8)

Warning: Values N>200 are approximate.  Plotting the power as a function
of N may reveal lower N values that have the required power.

napprox = 485

The result indicates that a power of 0.8 requires a sample size of 485. However, this result is approximate.

Make a plot to see if any smaller n values provide the required power of 0.8.

nn = 1:500;
pwrout = sampsizepwr('p',0.3,0.36,[],nn);
nexact = min(nn(pwrout>=0.8))

nexact = 462

figure
plot(nn,pwrout,'b-',[napprox nexact],pwrout([napprox nexact]),'ro')
grid on

The result indicates that a sample size of 462 also provides a power of 0.8 for this test.

A farmer wants to test the impact of two different types of fertilizer on the yield of his bean crops. He currently uses Fertilizer A, but believes that Fertilizer B might improve crop yield. Because Fertilizer B is more expensive than Fertilizer A, the farmer wants to limit the number of plants he treats with Fertilizer B in this experiment.

The farmer uses a 2:1 ratio of plants in each treatment group. He tests 10 plants with Fertilizer A, and 5 plants with Fertilizer B. The mean yield using Fertilizer A is 1.4 kg per plant, with a standard deviation of 0.2. The mean yield using Fertilizer B is 1.7 kg per plant. The significance level of the test is 0.05.

Compute the power of the test.

pwr = sampsizepwr('t2',[1.4 0.2],1.7,[],5,'Ratio',2)

pwr = 0.7165

The farmer wants to increase the power of the test to 0.90. Calculate how many plants he must treat with each type of fertilizer.

n = sampsizepwr('t2',[1.4 0.2],1.7,0.9,[])

n = 11

To increase the power of the test to 0.90, the farmer must test 11 plants with each type of fertilizer.

The farmer wants to reduce the number of plants he must treat with Fertilizer B, but keep the power of the test at 0.90 and maintain the initial 2:1 ratio of plants in each treatment group.

Using a 2:1 ratio of plants in each treatment group, calculate how many plants the farmer must test to obtain a power of 0.90. Use the mean and standard deviation values obtained in the previous test.

[n1out,n2out] = sampsizepwr('t2',[1.4,0.2],1.7,0.9,[],'Ratio',2)

n1out = 8

n2out = 16

To obtain a power of 0.90, the farmer must treat 16 plants with Fertilizer A and 8 plants with Fertilizer B.