Meat and Livestock Australia have commissioned your services to help with the design of an experiment to test whether using hormone growth promotant (HGP) drugs in cattle noticeably decreases the eating quality of grilled steaks.

Eating quality is assessed by untrained consumers in a sensory panel where they rate the product on a scale from 0 to 100.

Over the past 20 years of experimentation, we have developed a good understanding of the inherent variation in these consumer eating quality measurements. You can assume that the standard deviation is known to be 14.

Prior work in this field suggests that the use of HGP in cattle could reduce eating quality by 4 points.

Given the above information and using 0.05 as the level of significance, what is the statistical power of a two-sample t-test if 153 animals received the HGP treatment and 153 animals were HGP free?

Round your answer to 3 decimal places.

To calculate the statistical power of a two-sample t-test, we can use the following steps:

### Step 1: Gather the information
- **Sample size for each group (\( n_1 = n_2 = n \))**: 153
- **Standard deviation (\( \sigma \))**: 14
- **Mean difference to detect (\( \Delta \))**: 4 points
- **Level of significance (\( \alpha \))**: 0.05 (for a two-tailed test)
  
### Step 2: Calculate the standard error of the difference between means
\[
\text{Standard Error} = \sqrt{\frac{\sigma^2}{n_1} + \frac{\sigma^2}{n_2}} = \sqrt{\frac{14^2}{153} + \frac{14^2}{153}} = \sqrt{\frac{196}{153} + \frac{196}{153}} = \sqrt{\frac{392}{153}} = \sqrt{2.56} \approx 1.6
\]

### Step 3: Calculate the test statistic (\( z \)) under the null hypothesis
This is given by the mean difference divided by the standard error:
\[
z = \frac{\Delta}{\text{Standard Error}} = \frac{4}{1.6} = 2.5
\]

### Step 4: Determine the critical value for a two-tailed test at \( \alpha = 0.05 \)
The critical value \( z_{\alpha/2} \) for a significance level of 0.05 is 1.96 (from standard normal distribution tables).

### Step 5: Calculate the non-centrality parameter (\( \delta \))
\[
\delta = \frac{\Delta}{\text{Standard Error}} = 2.5
\]

### Step 6: Calculate the power
The power of the test is given by:
\[
\text{Power} = P\left(Z > z_{\alpha/2} - \delta\right) + P\left(Z < -z_{\alpha/2} - \delta\right)
\]

Using standard normal distribution tables or software to find the power corresponding to \( z = 2.5 \) and \( z_{\alpha/2} = 1.96 \):

The power is:
\[
\text{Power} = P\left(Z > 1.96 - 2.5\right) = P\left(Z > -0.54\right)
\]
\[
P(Z > -0.54) \approx 0.7054
\]

Thus, the power of the test is approximately 0.705 or **70.5%**.

**Answer:** The statistical power of the two-sample t-test is approximately **0.705**.

Would you like any further details or have any questions about the calculations? Here are some related questions:

1. How does increasing the sample size affect the power of the test?
2. What happens to the power if the expected mean difference changes?
3. How would the power change if the significance level (\( \alpha \)) is reduced?
4. What is the impact of variability (\( \sigma \)) on the power of a test?
5. How can we interpret a statistical power of 70.5% in practical terms?

**Tip:** Increasing the sample size or the effect size increases the power of a test, which reduces the probability of making a Type II error.

Learn how to calculate the statistical power of a two-sample t-test to assess the effect of hormone growth promotant (HGP) drugs on the eating quality of grilled steaks. Includes step-by-step calculations using known standard deviation, sample sizes, and expected mean difference.

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