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Refer to Exercise \(10.100 .\) Wishing to demonstrate that the variability of fills is less for her model than for her competitor's, a sales representative for company A acquired a sample of 30 fills from her company's model and a sample of 10 fills from her competitor's model. The sample variances were \(s_{A}^{2}=.027\) and \(s_{B}^{2}=.065,\) respectively. Does this result provide statistical support at the .05 level of significance for the sales representative's claim?

Short Answer

Expert verified
Answer: No, the test results do not provide enough evidence at the 0.05 level of significance to support the sales representative's claim that the variability of fills is less for her model than for the competitor's model.

Step by step solution

01

Compute the F-statistic

The F-statistic is the ratio of the larger sample variance to the smaller sample variance. In this case, we have: F = \(\frac{s_{A}^2}{s_{B}^2}\) F = \(\frac{0.027}{0.065}\) So, the F-statistic is: F = 0.415
02

Determine the F critical value

To determine the F critical value, we need the degrees of freedom for both samples. The degrees of freedom for the samples are \(df_A = n_A - 1 = 29\) and \(df_B = n_B - 1 = 9\). We will use the F-distribution table and a one-tailed test with a significance level of 0.05. Based on the degrees of freedom for both samples and the desired significance level, we can find the F critical value: F critical value = 2.24
03

Compare the F-statistic to the F critical value

We will now compare the computed F-statistic (0.415) to the F critical value (2.24) to determine if we can reject the null hypothesis. F-statistic = 0.415 F critical value = 2.24 Since the F-statistic is less than the F critical value, we cannot reject the null hypothesis.
04

Make a conclusion

The test results do not provide enough evidence at the 0.05 level of significance to support the sales representative's claim that the variability of fills is less for her model than for the competitor's model.

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Key Concepts

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

F-statistic
The F-statistic is an essential tool used in hypothesis testing when comparing two sample variances. It helps determine whether the variability of two data sets is significantly different.
In essence, the F-statistic is calculated by taking the ratio of two variances:
  • The variance of the sample with the larger variance (\( s_A^2 \)), and
  • The variance of the sample with the smaller variance (\( s_B^2 \)).
The formula is given by \( F = \frac{s_A^2}{s_B^2} \).
In our example, the sales representative wanted to compare the variance of fills from two different models.
The calculated F-statistic was 0.415. This ratio helps decide whether the observed difference between variances is large enough to conclude that one model indeed has less variability.
Degrees of Freedom
Degrees of freedom (df) are a crucial concept in statistics, representing the number of independent values that can vary in an analysis.
For sample variances, the degrees of freedom are calculated as the number of observations minus one.
  • Example: For company A's sample of 30 fills, the degrees of freedom are \( df_A = 30 - 1 = 29 \).
  • For company B's sample of 10 fills, the degrees of freedom are \( df_B = 10 - 1 = 9 \).
This concept helps determine the appropriate critical values from statistical distribution tables, like the F-distribution table in our exercise.
Sample Variance
Sample variance measures the spread or variability of data points within a particular sample.
It's calculated using the formula \( s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2 \), where \( n \) is the number of observations, \( x_i \) represents each observation, and \( \bar{x} \) is the sample mean.
  • Sample variance is essential in understanding how data is spread out, and it forms the basis of calculating the F-statistic.
  • In the problem, company A had a sample variance of 0.027, while company B had a sample variance of 0.065. These variances are used as indicators of variability in the fills from each company.
By comparing these sample variances using the F-statistic, we can assess if any significant differences exist between the two groups.
Significance Level
The significance level, often denoted by \( \alpha \), is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It reflects the probability of making a Type I error, which occurs when the null hypothesis is incorrectly rejected.
In most scientific studies, a common significance level used is 0.05, which translates to a 5% chance of making such an error.
  • In the exercise, a significance level of 0.05 was used, indicating that the sales representative would consider evidence significant if the probability of observing such data, assuming the null hypothesis is true, is less than 5%.
  • It helps when comparing the F-statistic to the critical F value from the F-distribution table. If the F-statistic is less than the critical value, the null hypothesis is not rejected, as in our example.
This ensures that the conclusion drawn is based on a predefined, acceptable level of risk.

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Most popular questions from this chapter

Jan Lindhe conducted a study on the effect of an oral antiplaque rinse on plaque buildup on teeth. \(^{6}\) Fourteen people whose teeth were thoroughly cleaned and polished were randomly assigned to two groups of seven subjects each. Both groups were assigned to use oral rinses (no brushing) for a 2 -week period. Group 1 used a rinse that contained an antiplaque agent. Group \(2,\) the control group, received a similar rinse except that, unknown to the subjects, the rinse contained no antiplaque agent. A plaque index \(x\), a measure of plaque buildup, was recorded at 14 days with means and standard deviations for the two groups shown in the table. $$ \begin{array}{lll} & \text { Control Group } & \text { Antiplaque Group } \\ \hline \text { Sample Size } & 7 & 7 \\ \text { Mean } & 1.26 & .78 \\ \text { Standard Deviation } & 32 & 32 \end{array} $$ a. State the null and alternative hypotheses that should be used to test the effectiveness of the antiplaque oral rinse b. Do the data provide sufficient evidence to indicate that the oral antiplaque rinse is effective? Test using \(\alpha=.05 .\) c. Find the approximate \(p\) -value for the test.

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