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The Gibbs Baby Food Company wishes to compare the weight gain of infants using their brand versus their competitor's. A sample of 40 babies using the Gibbs products revealed a mean weight gain of 7.6 pounds in the first three months after birth. The standard deviation of the sample was 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds, with a standard deviation of 2.9 pounds. At the .05 significance level, can we conclude that babies using the Gibbs brand gained less weight? Compute the \(p\) -value and interpret it.

Short Answer

Expert verified
Fail to reject the null hypothesis; Gibbs brand doesn't lead to significantly less weight gain.

Step by step solution

01

State the Hypotheses

Define the null and alternative hypotheses. Null Hypothesis, \( H_0 \): There is no difference in weight gain; babies using Gibbs brand do not gain less weight compared to the competitor. \( \mu_1 \geq \mu_2 \).Alternative Hypothesis, \( H_a \): Babies using Gibbs brand gain less weight compared to the competitor. \( \mu_1 < \mu_2 \).
02

Identify the Test Statistic

Since both sample sizes are greater than 30, we use a two-sample \( z \)-test for the difference in means. The test statistic is calculated as follows:\[z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]Where \( \bar{x}_1 = 7.6 \), \( s_1 = 2.3 \), \( n_1 = 40 \), \( \bar{x}_2 = 8.1 \), \( s_2 = 2.9 \), and \( n_2 = 55 \).
03

Compute the Test Statistic

Plug in the given values:\[z = \frac{(7.6 - 8.1) - 0}{\sqrt{\frac{2.3^2}{40} + \frac{2.9^2}{55}}}\]Calculate the standard error:\[\text{Standard Error} = \sqrt{\frac{5.29}{40} + \frac{8.41}{55}} = \sqrt{0.13225 + 0.15273} = \sqrt{0.28498} = 0.53408\]Then calculate the \( z \)-score:\[z = \frac{-0.5}{0.53408} \approx -0.936\]
04

Determine the Critical Value

Using the standard normal distribution table for \( \alpha = 0.05 \), the critical \( z \)-value for a one-tailed test is approximately \(-1.645\).
05

Find the p-value

Use the standard normal distribution to find the p-value corresponding to \( z = -0.936 \). The p-value is the probability of observing a test statistic as extreme as \(-0.936\).From the z-table, the p-value for \( z = -0.936 \) is approximately 0.1742.
06

Interpret the Results

Since the p-value (0.1742) is greater than the significance level (0.05), we fail to reject the null hypothesis. This means there is not enough evidence at the 5% level of significance to conclude that babies using the Gibbs brand gained less weight than those using the competitor's brand.

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Key Concepts

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

Two-Sample Z-Test
The two-sample z-test is a statistical tool designed to compare the means of two independent samples. In our case with the Gibbs Baby Food Company, we are interested in determining if there is a significant difference in weight gain between infants using their product versus a competitor's. This test is suitable here due to the large sample sizes (both greater than 30), which allows us to appropriately employ the z-test instead of a t-test.

The formula for the two-sample z-test is:
  • \[z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
Where:
  • \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means,
  • \( s_1 \) and \( s_2 \) are the standard deviations,
  • \( n_1 \) and \( n_2 \) are the sample sizes.
This method allows us to determine if the difference observed in the sample means is statistically significant or merely due to random sampling variability.
Significance Level
The significance level, often denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is actually true. It is essentially the threshold for the p-value that determines whether the results are statistically significant.

In the exercise, the significance level is set at 0.05 or 5%. This means the researchers accept a 5% chance of incorrectly concluding that Gibbs brand results in less weight gain when it actually does not. Choosing \( \alpha = 0.05 \) is common practice in many scientific studies, as it provides a balance between being too lenient, which could lead to false positives, and being too strict, which might ignore meaningful results.

When conducting the test, if the p-value is less than or equal to the significance level, the null hypothesis is rejected in favor of the alternative hypothesis. If it's greater, like in this study, the null hypothesis cannot be rejected.
Null and Alternative Hypotheses
The first critical step in hypothesis testing is defining the null and alternative hypotheses. These hypotheses form the foundation of statistical testing by clearly stating what is being tested and the assumptions involved.

In the Gibbs Baby Food example:
  • Null Hypothesis (\( H_0 \)): There is no difference in weight gain; babies using the Gibbs brand do not gain less weight compared to the competitor. Mathematically, this is represented as \( \mu_1 \geq \mu_2 \).
  • Alternative Hypothesis (\( H_a \)): Babies using the Gibbs brand gain less weight compared to the competitor. This is expressed as \( \mu_1 < \mu_2 \).
The null hypothesis is what the researcher seeks to disprove, while the alternative hypothesis is what the researcher is trying to provide evidence for. In hypothesis testing, we assess these hypotheses to decide whether to reject the null hypothesis.
P-Value Interpretation
Interpreting the p-value is central to understanding the results of a hypothesis test. A p-value indicates the probability of observing a test statistic as extreme as, or more extreme than, the actual observed results, assuming the null hypothesis is true.

In our analysis, we found a p-value of approximately 0.1742. This p-value tells us that there is a 17.42% chance of observing the sample results, or something more extreme, if the null hypothesis were true.

Since the p-value of 0.1742 is greater than our significance level of 0.05, it suggests that the observed difference in weight gain is not statistically significant. Consequently, we fail to reject the null hypothesis, indicating insufficient evidence to claim that babies using the Gibbs brand gained less weight than those using the competitor's brand.

P-values allow researchers to quantify the strength of evidence against the null hypothesis and aid in making informed decisions based on data.

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