Problem 17
Question
Experience raising New Jersey Red chickens revealed the mean weight of the chickens at five months is 4.35 pounds. The weights follow the normal distribution. In an effort to increase their weight, a special additive is added to the chicken feed. The subsequent weights of \(a\) sample of five-month- old chickens were (in pounds): $$\begin{array}{|lllllllll|}\hline 4.41 & 4.37 & 4.33 & 4.35 & 4.30 & 4.39 & 4.36 & 4.38 & 4.40 & 4.39 \\\\\hline\end{array}$$ At the .01 level, has the special additive increased the mean weight of the chickens? Estimate the \(p\) -value.
Step-by-Step Solution
Verified Answer
No, the additive has not significantly increased the mean weight.
1Step 1: State the Hypotheses
We need to test whether the special additive has increased the mean weight of the chickens. Our null hypothesis states that the mean weight with the additive is equal to the mean weight without it, \( H_0: \mu = 4.35 \). The alternative hypothesis states that the mean weight with the additive is greater, \( H_1: \mu > 4.35 \). This is a one-tailed test.
2Step 2: Calculate the Sample Mean
Calculate the mean of the sample to check the average weight of the chickens.Given sample weights: 4.41, 4.37, 4.33, 4.35, 4.30, 4.39, 4.36, 4.38, 4.40, 4.39.Formula for mean: \( \bar{x} = \frac{\sum{x}}{n} \) \[ \bar{x} = \frac{4.41 + 4.37 + 4.33 + 4.35 + 4.30 + 4.39 + 4.36 + 4.38 + 4.40 + 4.39}{10} = 4.368\]
3Step 3: Calculate the Standard Deviation
Determine the standard deviation of the sample as it will help us compute the test statistic.The sample variance \( s^2 \) is calculated as:\[ s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} \]After calculating:\[ s^2 = \frac{(4.41 - 4.368)^2 + ... + (4.39 - 4.368)^2}{9} = 0.00141\]Thus, the standard deviation \( s = \sqrt{0.00141} = 0.0376 \) pounds.
4Step 4: Calculate the Test Statistic
The test statistic for the sample mean is calculated as:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]Where \( \bar{x} = 4.368 \), \( \mu = 4.35 \), \( s = 0.0376 \), and \( n = 10 \).\[ t = \frac{4.368 - 4.35}{0.0376 / \sqrt{10}} = 1.52 \]
5Step 5: Determine the Critical Value and Decision
For \( \alpha = 0.01 \) and \( n-1 = 9 \) degrees of freedom, the critical value for a one-tailed test can be found from the t-distribution table as \( t_{critical} = 2.821 \).Since the calculated \( t \)-value (1.52) is less than the critical value (2.821), we fail to reject the null hypothesis.
6Step 6: Estimate the p-value
Using a t-distribution table or calculator, a \( t \)-value of 1.52 with 9 degrees of freedom does not reach significance at the 0.01 level and corresponds to a \( p \)-value greater than 0.10.Thus, the \( p \)-value is estimated to be \( p > 0.10 \).
Key Concepts
Understanding t-DistributionRole of Standard DeviationUnderstanding Null HypothesisEstimating the p-Value
Understanding t-Distribution
In hypothesis testing, the t-distribution is crucial when dealing with small sample sizes (typically n < 30) and when the population standard deviation is unknown. The t-distribution looks similar to the normal distribution but has heavier tails. This characteristic allows it to accommodate more variability when estimating the standard deviation from a sample rather than the entire population.
When using a t-distribution, the shape of the curve changes depending on the degrees of freedom, which are determined by the sample size (n-1). For example, with 9 degrees of freedom (our case here, as we have 10 samples), the t-distribution will have thicker tails compared to a normal distribution. This means it is more forgiving of data points that fall far from the mean.
The t-distribution plays a key role in calculating the test statistic in our exercise by helping us assess whether there is enough evidence to reject the null hypothesis.
When using a t-distribution, the shape of the curve changes depending on the degrees of freedom, which are determined by the sample size (n-1). For example, with 9 degrees of freedom (our case here, as we have 10 samples), the t-distribution will have thicker tails compared to a normal distribution. This means it is more forgiving of data points that fall far from the mean.
The t-distribution plays a key role in calculating the test statistic in our exercise by helping us assess whether there is enough evidence to reject the null hypothesis.
Role of Standard Deviation
Standard deviation measures the variation or spread of a set of values. In our exercise, calculating the standard deviation is vital because it tells us how much the chicken weights deviate from the mean weight, 4.35 pounds.
The formula for standard deviation, derived from the variance, is \( s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} \). Here, \( x_i \) represents each individual weight of the chickens, \( \bar{x} \) is the sample mean, and \( n-1 \) accounts for degrees of freedom.
Once calculated, the standard deviation is used in the t-statistic formula to normalize our sample mean's deviation from the hypothesized population mean. A smaller standard deviation signifies data that's close to the mean, while a larger one indicates more spread out data.
The formula for standard deviation, derived from the variance, is \( s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} \). Here, \( x_i \) represents each individual weight of the chickens, \( \bar{x} \) is the sample mean, and \( n-1 \) accounts for degrees of freedom.
Once calculated, the standard deviation is used in the t-statistic formula to normalize our sample mean's deviation from the hypothesized population mean. A smaller standard deviation signifies data that's close to the mean, while a larger one indicates more spread out data.
Understanding Null Hypothesis
In statistics, the null hypothesis is a statement that suggests no effect or no difference exists in a particular phenomenon. It serves as the baseline or default assumption that any observed variation is due to random chance rather than a real effect.
For this exercise, the null hypothesis \( H_0: \mu = 4.35 \) asserts that the mean weight of chickens remains unchanged after adding the feed additive. Our goal in testing is to determine whether collected sample data provides enough evidence to reject this assumption.
If the test statistic (found using our sample data) exceeds the critical value from the t-table, it implies a significant difference, letting us reject the null hypothesis. Otherwise, we "fail to reject" it, meaning insufficient evidence to claim a difference. The key lies in the "fail to reject" term, because failing to reject doesn't prove the null hypothesis; it just lacks evidence against it.
For this exercise, the null hypothesis \( H_0: \mu = 4.35 \) asserts that the mean weight of chickens remains unchanged after adding the feed additive. Our goal in testing is to determine whether collected sample data provides enough evidence to reject this assumption.
If the test statistic (found using our sample data) exceeds the critical value from the t-table, it implies a significant difference, letting us reject the null hypothesis. Otherwise, we "fail to reject" it, meaning insufficient evidence to claim a difference. The key lies in the "fail to reject" term, because failing to reject doesn't prove the null hypothesis; it just lacks evidence against it.
Estimating the p-Value
The p-value is a probability measure that helps us determine the significance of our test results. It gives us the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true.
In the context of our exercise, we calculated a t-value and compared it with the critical t-value from the t-distribution table. However, rather than relying on just a threshold, the p-value provides a more nuanced view.
A small p-value (typically less than \( \alpha = 0.01\) for our test) suggests that the observed data is unlikely under the null hypothesis, leading us to reject it. Here, since the calculated p-value was greater than 0.10, it implies insufficient evidence to reject the null hypothesis, and thus the special additive was unlikely to have significantly increased the chicken's weight.
In the context of our exercise, we calculated a t-value and compared it with the critical t-value from the t-distribution table. However, rather than relying on just a threshold, the p-value provides a more nuanced view.
A small p-value (typically less than \( \alpha = 0.01\) for our test) suggests that the observed data is unlikely under the null hypothesis, leading us to reject it. Here, since the calculated p-value was greater than 0.10, it implies insufficient evidence to reject the null hypothesis, and thus the special additive was unlikely to have significantly increased the chicken's weight.
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