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Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its 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. For the Gibbs brand the population standard deviation of the sample is 2.3 pounds. A sample of 55 babies using the competitor's brand revealed a mean increase in weight of 8.1 pounds. The population standard deviation is 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
No, we cannot conclude Gibbs babies gained less weight; p-value is 0.1797.

Step by step solution

01

Define the Hypotheses

We need to test if the weight gain for babies using the Gibbs brand is less than those using the competitor's brand. Therefore, we set up the null hypothesis as \( H_0: \mu_G = \mu_C \) (mean weight gain of Gibbs equals competitor) and the alternative hypothesis as \( H_a: \mu_G < \mu_C \) (mean weight gain of Gibbs is less than competitor).
02

Gather and Note Information

For Gibbs: sample size \( n_1 = 40 \), mean \( \bar{x}_1 = 7.6 \), standard deviation \( \sigma_1 = 2.3 \). For Competitor: sample size \( n_2 = 55 \), mean \( \bar{x}_2 = 8.1 \), standard deviation \( \sigma_2 = 2.9 \).
03

Calculate the Test Statistic

The test statistic for comparing two means with known population standard deviations is given by: \[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Plugging in the values gives: \[z = \frac{7.6 - 8.1}{\sqrt{\frac{2.3^2}{40} + \frac{2.9^2}{55}}} \approx -0.916\]
04

Find the P-value

Using the standard normal distribution table, we find that the p-value for \( z = -0.916 \) is approximately 0.1797.
05

Conclusion and Interpretation

At the \( \alpha = 0.05 \) significance level, compare the p-value to \( \alpha \). Since 0.1797 > 0.05, we fail to reject the null hypothesis. This means there is not enough statistical evidence to support 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.

Statistical Significance
Statistical significance is a measure that helps us determine whether the results of a study are meaningful enough to support a specific hypothesis. In the context of hypothesis testing, this usually involves comparing a calculated value (like a p-value) to a significance level (often set at 0.05).
In our scenario, the researchers want to know if infants using Gibbs Baby Food gained significantly less weight compared to those using a competitor's brand. To do this, they set a significance level, or alpha (α), at 0.05. This means they are willing to accept a 5% chance of incorrectly rejecting a true null hypothesis. The null hypothesis here is that the weight gain from using Gibbs is the same as using the competitor's brand.
  • If the p-value is less than 0.05, we would reject the null hypothesis, showing statistical significance. In other words, the difference in weight gain would be considered not due to random chance.
  • However, if the p-value is greater than 0.05, as it is in this case (0.1797), we fail to reject the null hypothesis.
This tells us that the observed difference in weight gain is likely due to chance and not statistically significant at the 0.05 level.
P-Value Interpretation
The p-value is a crucial element in hypothesis testing as it helps summarize the evidence against the null hypothesis. It tells us how likely it is to observe a result as extreme as, or more extreme than, the one in our study, assuming that the null hypothesis is true.
In our exercise, the p-value is 0.1797, which was calculated by using the standard normal distribution and a z-score of -0.916. Here's how you interpret it:
- A small p-value (typically less than 0.05) would indicate strong evidence against the null hypothesis, leading us to conclude that the difference in weight gain is significant. - A p-value of 0.1797 is not small enough; thus, it suggests that there is not sufficient evidence to suggest that the difference in the weight gains is significant. So, in simpler terms, since our p-value is greater than 0.05, we continue to assume that there is no difference in the weight gains between the two brands. This interpretation helps us make informed decisions about our hypotheses.
Standard Normal Distribution
The standard normal distribution is a key concept in statistics, especially when dealing with hypothesis testing and p-values. It's a special type of normal distribution where the mean is 0 and the standard deviation is 1.
In hypothesis testing, we often use the standard normal distribution to find probabilities and critical values when comparing sample statistics to population parameters. For tests involving means, like in our study comparing weight gains between two baby food brands, the calculated test statistic (z-score) often refers back to this distribution.
- Here, the standard normal distribution was used to compute a z-score of approximately -0.916. This score measures how many standard deviations away our sample mean (the difference between Gibbs and the competitor's mean weight gains) is from the null value (assumed to be 0). - A standard normal distribution table (or an equivalent calculator) converts the z-score into a p-value, helping us to assess statistical significance.
Understanding this concept is crucial because it underpins the determination of p-values and the subsequent decision-making regarding statistical significance in hypothesis testing.

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