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A t-test is a statistical test used to compare the means of two groups. When you're comparing a sample mean to a known population mean, you're often testing to see if there is a significant difference between them. In our chicken weight example, the t-test tells us if the special additive in chicken feed causes a meaningful difference in the mean weight compared to the established mean of 4.35 pounds.

The t-test relies on a few key assumptions:

• The data should be approximately normally distributed.
• The sample should be a simple random sample from the population.
• The variance of the sample should be similar to that of the population.

In this exercise, the decision is based on the calculated t-value. If this value is greater than the critical t-value, the difference in means is statistically significant. If not, we conclude that there's no sufficient evidence to suggest that the special additive has significantly increased the weight of the chickens.

The sample mean, denoted by \( \bar{x} \), is the average weight of chickens in our sample. It's calculated by adding all the sample weights together and dividing by the number of observations.

In this problem, we computed the sample mean as follows: \[ \bar{x} = \frac{4.41 + 4.37 + 4.33 + 4.35 + 4.30 + 4.39 + 4.36 + 4.38 + 4.40 + 4.39}{10} = 4.368 \] This sample mean is essential because it provides a point of comparison against the known population mean (4.35 pounds). By conducting a t-test, we determine if this sample mean is significantly greater than the population mean, suggesting that the additive has had an effect.

Standard deviation, represented by \( s \), measures how spread out the sample weights are around the sample mean. It helps determine the variability within a sample.

To find the standard deviation, you first calculate the variance \( s^2 \), and then take the square root of that variance. In our example, we calculated the variance: \[ s^2 = \frac{(4.41 - 4.368)^2 + (4.37 - 4.368)^2 + \cdots + (4.39 - 4.368)^2}{10 - 1} \approx 0.00245 \] and then found the standard deviation to be: \[ s = \sqrt{0.00245} \approx 0.0495 \] This value shows us how much individual chicken weights vary from the mean weight. A low standard deviation indicates the weights are closely clustered around the mean, while a high standard deviation indicates more spread.

The p-value is a probability measure that helps us decide whether to reject the null hypothesis. It tells us the likelihood of observing a test statistic as extreme as the one computed, assuming the null hypothesis is true.

A low p-value means the observed data is very unlikely under the null hypothesis, suggesting strong evidence against it. In our exercise, we estimate the p-value using a t-distribution table. Since our calculated t-value of 1.152 is less than the critical t-value of 2.821 at the 0.01 significance level, the p-value is greater than 0.01, indicating insufficient evidence to reject the null hypothesis.

When evaluating p-values:

• If the p-value is less than or equal to the chosen significance level (e.g., \( \alpha = 0.01 \)), reject the null hypothesis.
• If the p-value is greater, we do not reject the null hypothesis.

This careful evaluation lets us conclude whether the special additive truly affects the chicken weights.