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The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the deflection temperatures observed are as follows (in \({ }^{\circ} \mathrm{F}\) ): $$\begin{array}{l}\text { Type } 1: 206,188,205,187,194,193,207,185,189,213,192,210,194,178,205 \end{array}$$ $$\begin{array}{l}\text { Type } 2: 177,197,206,201,180,176,185,200,197,192, 198,188,189,203,192 \end{array}$$ (a) Construct box plots and normal probability plots for the two samples. Do these plots provide support of the assumptions of normality and equal variances? Write a practical interpretation for these plots. (b) Do the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type \(2 ?\) In reaching your conclusions, use \(\alpha=0.05 .\) Calculate a \(P\) -value (c) If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as \(5^{\circ} \mathrm{F}\), it is important to detect this difference with probability at least \(0.90 .\) Is the choice of \(n_{1}=n_{2}=15\) adequate? Use \(\alpha=0.05\)

Step by step solution

01

Construct Box Plots

To construct box plots, first organize the data for both types in ascending order. Plot these as box plots for each sample: - **Type 1**: Average temperature is centered around the box, with quartiles indicating symmetry or skewness. - **Type 2**: Similarly, the box plot will show central tendency and dispersion. Look for overlaps and spread to gauge equal variances. Evaluate skewness and outliers for normality. Ideally, box plots should be symmetric without outliers to assume normality.

02

Construct Normal Probability Plots

Create normal probability plots for the data in each type. Plot each data point against a theoretical normal distribution. - **Type 1** and **Type 2** should show a roughly straight line if the data follows a normal distribution. Deviations from this line could imply non-normality.

03

Interpret Box and Probability Plots

Compare spread and shapes of box plots to check variances. Use normal probability plots to assess normality. - Equal spread in box plots implies equal variances. - A linear pattern in probability plots indicates normality. Summarize if assumptions hold based on plots.

04

Formulate Hypothesis for Temperature Comparison

To test if the deflection temperature of type 1 exceeds type 2: - **Null Hypothesis (H鈧�)**: No difference in means between type 1 and type 2. - **Alternative Hypothesis (H鈧�)**: Mean temperature of type 1 > type 2.

05

Conduct Two-Sample t-test

Assuming equal variances, use the t-test formula:\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{2}{n}}}\]Where:- \(\bar{x}_1, \bar{x}_2\) are means of type 1 and 2.- \(s_p\) is the pooled standard deviation.- \(n\) is the sample size. Calculate the t-statistic and p-value.

06

Compare P-value to Significance Level

Evaluate the p-value under \(\alpha = 0.05\). If p-value < 0.05, reject H鈧� in favor of H鈧�, supporting that type 1 exceeds type 2.Summarize decision from test evidence.

07

Calculate Power of Test for Detectable Difference

To determine if sample size is adequate, calculate the power of the test:- Set detectable difference \( \delta = 5 \)- Calculate required power using:\[1 - \beta = P(Z > z_\alpha + \frac{\delta}{s_p\sqrt{2/n}})\]where \( Z \) is the standard normal deviate.Check if power > 0.90, assessing sample adequacy.

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Key Concepts

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

Box Plot

Box plots are a great way to visually display the distribution of data and summarize its key characteristics like central tendency and variability. For Type 1 and Type 2 plastic pipes, you can create box plots by first organizing each set of data in ascending order. The box plot will show you the median (center line of the box), the quartiles (edges of the box), and possible outliers (points outside the whiskers).

When you look at these plots, pay attention to:

• Symmetry - If the box plot appears roughly symmetrical, this suggests that the data might be normally distributed.
• Spread - The length of the box indicates the interquartile range, which helps in assessing the variance of the data.
• Outliers - Points that lie outside the whiskers could indicate anomalies or deviations from normality.

By comparing the box plots for Type 1 and Type 2, if they appear similar in spread and symmetry, this supports the assumption of equal variances.

Normal Probability Plot

A normal probability plot, also known as a Q-Q plot, is useful in determining whether a data set follows a normal distribution. To create one, each data point is plotted against the expected value from a normal distribution.

Here's what you should look for:

• A Straight Line - If the data points form a roughly straight line, this indicates that the data approximates a normal distribution.
• Curvature or Deviation - Any noticeable curvature or deviation from the line could imply the data is not normal.

For both types of plastic pipes, drawing such plots helps validate the assumption of normality, which is crucial for further statistical analyses, such as the t-test. When evaluating, ensure both Type 1 and Type 2 are aligned along a straight line to support the normality assumption.

Two-Sample t-test

The two-sample t-test is a statistical method used to determine if there is a significant difference between the means of two groups. In this case, you assess whether the mean deflection temperature of Type 1 pipes exceeds that of Type 2.

To conduct a two-sample t-test:

• Formulate Hypotheses:
  - Null Hypothesis ( H鈧�): The means of both types are equal.
  - Alternative Hypothesis ( H鈧�): The mean of Type 1 is greater than Type 2.
• Assume equal variances, if supported by plots.
  
• Calculate the t-statistic using:\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{2}{n}}}\]where \( \bar{x}_1 \) and \( \bar{x}_2 \) are the sample means, \( s_p \) is the pooled standard deviation, and \( n \) is the sample size.
• Determine the p-value - If the p-value is below 0.05, you reject the null hypothesis, supporting that Type 1 exceeds Type 2.

This statistical test helps in making a probabilistic decision about the differences in mean temperatures based on sample data.

Statistical Power

Statistical power is the probability that a test will correctly reject a false null hypothesis. In simpler terms, it's the likelihood of detecting a true effect if there really is one. For the plastic pipe temperature tests, the desired power is at least 0.90, meaning there's a 90% chance of detecting a 5 degree difference if it exists.

To assess power adequacy:

• Calculate the power using the formula:\[1 - \beta = P(Z > z_\alpha + \frac{\delta}{s_p\sqrt{2/n}})\]where \( \delta \) is the detectable difference (here, 5 degrees), and \( Z \) is the standard normal deviate.
• Check if the calculated power meets or exceeds 0.90.

If the power is insufficient, you may need to increase the sample size to reliably detect the temperature difference, ensuring your conclusions are statistically robust.