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Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but you suspect that the variability in concentration may differ for the two companies. The standard deviation of concentration in a random sample of \(n_{1}=10\) batches produced by company 1 is \(s_{1}=4.7\) grams per liter, and for company \(2,\) a random sample of \(n_{2}=16\) batches yields \(s_{2}=5.8\) grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use \(\alpha=0.05\).

Short Answer

Expert verified
No, there is not enough evidence to conclude that the variances are different.

Step by step solution

01

Define the hypothesis

We want to test whether the variances of the concentrations from the two companies differ. This is a two-tailed test for equality of variances. The hypotheses are: - Null Hypothesis, \( H_0 \): \( \sigma_1^2 = \sigma_2^2 \)- Alternative Hypothesis, \( H_a \): \( \sigma_1^2 eq \sigma_2^2 \)
02

Determine the test statistic

We'll use the F-test to compare two variances:\[ F = \frac{s_1^2}{s_2^2} \]where \(s_1=4.7\) and \(s_2=5.8\). Compute \( s_1^2 \) and \( s_2^2 \):- \( s_1^2 = (4.7)^2 = 22.09 \)- \( s_2^2 = (5.8)^2 = 33.64 \)Thus, the test statistic is:\[ F = \frac{22.09}{33.64} \approx 0.656 \]
03

Determine critical values

The degrees of freedom for the numerator is \( df_1 = n_1 - 1 = 9 \) and for the denominator is \( df_2 = n_2 - 1 = 15 \). Using an F-distribution table and \( \alpha = 0.05 \), determine the critical values for a two-tailed test:- Upper critical value \( F_{0.025, 9, 15} \approx 3.27 \)- Lower critical value is the reciprocal: \( \frac{1}{3.27} \approx 0.306 \)
04

Make a decision

Compare the calculated \( F \) to the critical values:- If \( F < 0.306 \) or \( F > 3.27 \), reject the null hypothesis.- Our calculated \( F = 0.656 \) is not less than 0.306 and not greater than 3.27.Thus, we do not reject \( H_0 \).
05

Conclusion

There is not enough evidence at the \( \alpha = 0.05 \) significance level to conclude that the variances of concentrations from both companies are different.

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Key Concepts

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

F-test
The F-test is a statistical method used to compare two variances. It is often employed when one wants to determine if there is a significant difference between the variability, or spread, in two different data sets. In our case, we're using the F-test to examine if the variance in the concentration of a substance differs between two suppliers.
The F-test calculates an F-ratio, which takes the form of a ratio of two sample variances. This ratio allows us to assess whether there are any differences in variability between the two groups. The formula is given by:
\[ F = \frac{s_1^2}{s_2^2} \]
where \( s_1 \) and \( s_2 \) are the standard deviations of the two samples.
  • If the F-ratio is close to 1, it implies that the variances are similar.
  • If the F-ratio is significantly different from 1, it might indicate a difference in variances.
To make this determination, the calculated F-ratio is compared to a critical value obtained from the F-distribution table based on the chosen significance level \( \alpha \).
Variance Comparison
Variance comparison is an essential part of statistical analysis. It helps us understand how spread out or "varied" data points are in a given set of values. Like standard deviation, variance provides a measure of the degree to which each number in a data set differs from the mean of the dataset.
In this exercise, we are specifically comparing the variance in concentration from two different companies. The question at hand is whether these variances are significantly different, which would imply different consistency levels in materials these companies supply.
To compare variances effectively, we use the F-test. Given two different sample variances \(s_1^2\) and \(s_2^2\), we calculate the F-statistic and compare it to critical values. By assessing if the ratio is far from 1, we can make informed decisions about the similarity or difference in variances across the two samples.
Two-tailed test
A two-tailed test in statistics examines if a sample is significantly greater or significantly less than a population parameter. It's called "two-tailed" because we are checking for changes in both directions – higher and lower.
This type of test is typically used when researchers are interested in identifying any statistically significant difference between two populations without specifying the direction of the difference.
In our scenario, we're testing whether there is a difference in variances without assuming beforehand that one will be higher or lower. The null hypothesis assumes no difference, while the alternative hypothesis checks for inequality (either greater or lesser).
The two-tailed test uses critical values from the F-distribution on both ends of the distribution curve:
  • If the calculated test statistic falls into the critical "rejection" regions on either tail, we reject the null hypothesis.
  • Otherwise, we do not have enough evidence to conclude a variance difference.
This method ensures comprehensive testing for variability differences by covering both potential possibilities.
Null Hypothesis
The null hypothesis, often denoted as \( H_0 \), is a foundational concept in statistical hypothesis testing. It represents a default position or assumption that there is no effect or no difference.
When we perform hypothesis testing, our primary goal is to test the validity of the null hypothesis against an alternative hypothesis (\( H_a \)). The alternative hypothesis posits that there is a significant effect or difference.
In our exercise, the null hypothesis states that the variances in concentration from the two companies are equal, represented as:
\[ H_0: \sigma_1^2 = \sigma_2^2 \]
By default, we begin with the assumption that the null hypothesis is true. We then calculate the relevant test statistic (in this case, the F-statistic) to determine if there is enough evidence to reject it. If the test statistic falls within the critical values range, we cannot reject the null hypothesis. Otherwise, we would reject \( H_0 \) in favor of the alternative hypothesis.
The null hypothesis is crucial as it helps in maintaining a scientific approach to testing, ensuring that any conclusions drawn are statistically substantiated.

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Most popular questions from this chapter

The diameter of steel rods manufactured on two different extrusion machines is being investigated. Two random samples of sizes \(n_{1}=15\) and \(n_{2}=17\) are selected, and the sample means and sample variances are \(\bar{x}_{1}=8.73, s_{1}^{2}=\) \(0.35, \bar{x}_{2}=8.68,\) and \(s_{2}^{2}=0.40,\) respectively. Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from a normal distribution. a. Is there evidence to support the claim that the two machines produce rods with different mean diameters? Use \(\alpha=0.05\) in arriving at this conclusion. Find the \(P\) -value. b. Construct a \(95 \%\) confidence interval for the difference in mean rod diameter. Interpret this interval.

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