91Ó°ÊÓ

In an investigation of the toxin produced by a certain poisonous snake, a researcher prepared 26 different vials, each containing \(1 \mathrm{~g}\) of the toxin, and then determined the amount of antitoxin needed to neutralize the toxin. The sample average amount of antitoxin necessary was found to be \(1.89 \mathrm{mg}\), and the sample standard deviation was .42. Previous research had indicated that the true average neutralizing amount was \(1.75 \mathrm{mg} / \mathrm{g}\) of toxin. Does the new data contradict the value suggested by prior research? Test the relevant hypotheses using the \(P\)-value approach. Does the validity of your analysis depend on any assumptions about the population distribution of neutralizing amount? Explain.

Step by step solution

01

Define Hypotheses

Start by defining the null and alternative hypotheses. The null hypothesis \( H_0 \) states that the true average neutralizing amount is 1.75 mg/g, while the alternative hypothesis \( H_a \) suggests it is not equal to 1.75 mg/g, which indicates a two-tailed test. Thus, \( H_0: \mu = 1.75 \) and \( H_a: \mu eq 1.75 \).

02

Calculate the Test Statistic

Use the formula for the test statistic for a sample mean: \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} = 1.89 \), \( \mu = 1.75 \), \( s = 0.42 \), and \( n = 26 \). Substituting the values gives \( t = \frac{1.89 - 1.75}{0.42/\sqrt{26}} \approx 1.685 \).

03

Determine the Significance Level

Choose a significance level \( \alpha \). Traditionally, \( \alpha = 0.05 \) is used for a standard test. This value will be used to compare against the \( P \)-value to decide whether to reject the null hypothesis.

04

Find the P-value

Using a t-distribution table or calculator with \( n-1 = 25 \) degrees of freedom, calculate the \( P \)-value for \( t = 1.685 \). The \( P \)-value represents the probability of observing a test statistic at least as extreme as the one calculated.

05

Compare P-value to Alpha

If the \( P \)-value is less than \( \alpha \), reject the null hypothesis. For \( t = 1.685 \), the \( P \)-value is approximately 0.105. Since 0.105 is greater than 0.05, fail to reject the null hypothesis.

06

Consider Assumptions

The validity of the analysis assumes that the sample is drawn from a normally distributed population, or that the sample size is large enough (typically \( n > 30 \)) for the central limit theorem to apply. Given the sample size is 26, checking normality might be necessary for more rigorous analysis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

T-distribution

The T-distribution is a fundamental concept in statistics, especially when working with small sample sizes. It's similar to the normal distribution but has heavier tails, meaning it provides a better estimation when we have less information about the population.

• When we calculate a test statistic for a sample mean, like in our snake toxin example, the sample size is crucial. Here, we have a sample size of 26, which isn't very large.
• For such smaller samples, we use the T-distribution instead of the normal distribution to account for the additional uncertainty present when estimating population parameters.
• The shape of the T-distribution depends on the degrees of freedom (df). In our case, df = 25, which is one less than the sample size.

Because of its broader tails, the T-distribution allows us to make more cautious claims about statistical significance, helping us avoid false positives in hypothesis testing.

Null Hypothesis

The null hypothesis, often denoted as \( H_0 \), is a statement that assumes no effect or no difference. It's a starting point for testing our assumptions.

• In our exercise, the null hypothesis is \( H_0: \mu = 1.75 \). This suggests that the known average amount of antitoxin required (based on previous research) is accurate.
• The null hypothesis acts as a baseline assertion that any observed variance in our data is due to random chance, rather than a real effect or difference.
• Hypothesis testing allows us to challenge \( H_0 \) and determine whether we have enough evidence to support an alternative perspective.

If our analysis shows that the data significantly deviates from what the null hypothesis describes, we may reject it, indicating that the new findings might be true.

Sample Mean

The sample mean is a critical statistic in hypothesis testing. It provides a point estimate of the population mean.

• Here, the sample mean \( \bar{x} \) is 1.89 mg/g. This represents the average antitoxin amount calculated from the 26 vials.
• The sample mean is used to compare against the hypothesized population mean (1.75 mg/g in our case) to determine if any significant difference exists.
• A key component in calculating the test statistic in hypothesis testing, the sample mean indicates whether observed data align with the assumptions made under the null hypothesis.

Together with the sample standard deviation and size, it helps in shaping the argument for or against the null hypothesis, especially through the lens of T-distributions.

P-value

The P-value is a probabilistic metric, critical in decision-making during hypothesis testing.

• It measures the strength of the evidence against the null hypothesis, given the observed data.
• In our analysis, the calculated P-value for \( t = 1.685 \) is 0.105. This value indicates the probability of observing such a test statistic given that the null hypothesis is true.
• A smaller P-value suggests stronger evidence against \( H_0 \), often leading to rejection. Conversely, a larger P-value (such as ours, 0.105, which is above the common alpha level of 0.05) suggests insufficient evidence to reject \( H_0 \).
• Thus, P-values help to quantify the uncertainty, assisting researchers in determining whether deviations from \( H_0 \) are meaningful or mere statistical fluctuation.

Understanding P-values empowers decision-making in statistical inferences, guiding whether to maintain or discard the null hypothesis.