91Ó°ÊÓ

An article in Polymer Degradation and Stability \((2006,\) Vol. 91) presented data from a nine-year aging study on S537 foam. Foam samples were compressed to \(50 \%\) of their original thickness and stored at different temperatures for nine years. At the start of the experiment as well as during each year, sample thickness was measured, and the thicknesses of the eight samples at each storage condition were recorded. The data for two storage conditions follow. $$\begin{array}{ll}50^{\circ} \mathrm{C} & 0.047,0.060,0.061,0.064,0.080,0.090,0.118 \\\& 0.165,0.183 \\\60^{\circ} \mathrm{C} &0.062,0.105,0.118,0.137,0.153,0.197,0.210 \\\& 0.250,0.375\end{array}$$ (a) Is there evidence to support the claim that mean compression increases with the temperature at the storage condition? (b) Find a \(95 \%\) confidence interval for the difference in the mean compression for the two temperatures. (c) Is the value zero contained in the \(95 \%\) confidence interval? Explain the connection with the conclusion you reached in part (a). (d) Do normal probability plots of compression indicate any violations of the assumptions for the tests and confidence interval that you performed?

Step by step solution

01

Calculate Sample Means

Firstly, calculate the mean of the compression data for each temperature. For 50°C, add up all the measurements and divide by 9. Similarly, find the mean for 60°C using the same method.

02

Calculate Sample Variances

Find the variance for each sample set. To do this, subtract the mean from each data point, square the result, sum these squares, and then divide by the number of observations minus one. Repeat this for both temperatures.

03

Formulate Hypothesis for Mean Comparison

The hypothesis for part (a) is: - Null Hypothesis ( H_0 ): The mean compression at 50°C is equal to the mean compression at 60°C. - Alternative Hypothesis ( H_a ): The mean compression at 60°C is greater than at 50°C.

04

Perform t-Test for Equal Means

Perform a t-test using the means, variances, and sample sizes from Step 1 and Step 2. Determine whether the results provide evidence against the null hypothesis in favor of the alternative hypothesis.

05

Calculate 95% Confidence Interval

Use the sample means and variances from Step 1 and Step 2 to compute the 95% confidence interval for the difference in means. Use the formula for the confidence interval for a difference in means (with unequal variances).

06

Check if Zero is in the Confidence Interval

For part (c), examine whether zero is within the interval from Step 5. If the interval does not contain zero, it suggests a significant difference in compressions.

07

Assess Normality with Probability Plots

Construct normal probability plots for the data from each temperature condition. Assess if the data reasonably fit a straight line, indicating that the normality assumption holds.

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Key Concepts

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

t-test

A t-test is a statistical test used to compare the means of two groups. In our exercise, a t-test helps determine whether the average foam compression at different temperatures, 50°C and 60°C, significantly differs.

• Calculate the mean for each group. This represents the average compression.
• Determine the variance for each group, indicating the spread of the compression measurements.
• The t-test evaluates the null hypothesis (that there's no difference in means) against the alternative hypothesis (that the mean compression is different).

The outcome of the t-test shows if temperature significantly impacts foam compression.

confidence interval

A confidence interval gives us a range within which we expect the true mean difference to lie with a certain level of confidence, typically 95%.

• Compute the difference between sample means from each temperature group.
• Use the variance data to estimate the standard error of the mean difference.
• Apply the formula for the confidence interval to find the range of plausible mean differences.

If the interval excludes zero, it suggests a significant difference, implying one mean is likely greater than the other in the population.

normal probability plot

A normal probability plot helps check if data aligns with a normal distribution. This is crucial because many statistical tests, like the t-test, assume normality.

• Plot the data quantiles against a theoretical normal distribution's quantiles.
• If the plot resembles a straight line, the data is approximately normal.
• Any deviation may suggest skewness or other distribution abnormalities.

For our compression data, examining these plots confirms whether normality holds, validating our statistical tests.

hypothesis testing

Hypothesis testing is a method to decide if there's enough evidence to reject a stated hypothesis, typically the null hypothesis.

• Formulate hypotheses: the null hypothesis presumes no effect, while the alternative suggests a significant effect.
• Perform a test, like the t-test, to compare means or other parameters of interest.
• Assess the p-value to determine statistical significance. A low p-value (typically < 0.05) indicates strong evidence against the null hypothesis.

In our exercise, hypothesis testing reveals whether temperature significantly influences foam's compression over time.