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Two types of plastic are suitable for use by an electronics component manufacturer. The breaking strength of this plastic is important. It is known that \(\sigma_{1}=\sigma_{2}=1.0\) psi. From a random sample of size \(n_{1}=10\) and \(n_{2}=12,\) we obtain \(\bar{x}_{1}=162.5\) and \(\bar{x}_{2}=155.0 .\) The company will not adopt plastic 1 unless its mean breaking strength exceeds that of plastic 2 by at least 10 psi. (a) Based on the sample information, should it use plastic \(1 ?\) Use \(\alpha=0.05\) in reaching a decision. Find the \(P\) -value. (b) Calculate a \(95 \%\) confidence interval on the difference in means. Suppose that the true difference in means is really 12 psi. (c) Find the power of the test assuming that \(\alpha=0.05\). (d) If it is really important to detect a difference of 12 psi, are the sample sizes employed in part (a) adequate, in your opinion?

Short Answer

Expert verified
Plastic 1 should not be used as the initial difference is less than required. The 95% CI doesn't exclude 10 psi, the power is low, and sample sizes are inadequate.

Step by step solution

01

Define Hypotheses

We need to define the null and alternative hypotheses. The null hypothesis (H_0) is that the difference in mean breaking strengths for plastic 1 and plastic 2 is 10 psi: \(H_0: \mu_1 - \mu_2 = 10\). The alternative hypothesis (H_a) is that plastic 1 has a mean breaking strength that exceeds plastic 2 by more than 10 psi: \(H_a: \mu_1 - \mu_2 > 10\).
02

Compute Test Statistic

We use the formula for the test statistic for two independent means, given by:\(z = \frac{\bar{x}_1 - \bar{x}_2 - 10}{\sqrt{\left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right)}}\). Substituting the given values: \(\bar{x}_1 = 162.5\), \(\bar{x}_2 = 155.0\), \(\sigma_1 = 1.0\), \(\sigma_2 = 1.0\), \(n_1 = 10\), \(n_2 = 12\), we find \(z = \frac{162.5 - 155.0 - 10}{\sqrt{\frac{1.0^2}{10} + \frac{1.0^2}{12}}}\). Calculate this to find the test statistic value.
03

Find Critical Value and P-Value

With \(\alpha = 0.05\) for a one-tailed test, we look up the critical z-value which is 1.645. We compare the test statistic calculated in step 2 with this critical value. Additionally, calculate the p-value from the z-score to see the exact significance level of the result.
04

Comparison of Test Statistic with Critical Value

If the test statistic from step 2 exceeds the critical value of 1.645, we reject the null hypothesis. Otherwise, we do not reject it. This determines whether to use plastic 1.
05

Calculate 95% Confidence Interval

To find a 95% confidence interval for difference in means, use the formula: \[CI = (\bar{x}_1 - \bar{x}_2) \pm z_{(1-\alpha/2)} \cdot \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\].Substitute: \(\bar{x}_1 = 162.5\), \(\bar{x}_2 = 155.0\), and \(z_{0.975} = 1.96\) to calculate the interval.
06

Power of the Test

The power of the test is the probability of correctly rejecting the null hypothesis, given that a difference of 12 psi truly exists. It is calculated using the non-centrality parameter: \(\Delta = \frac{12}{\sqrt{\frac{1^2}{10} + \frac{1^2}{12}}}\). Using the distribution of the test statistic, find the power by calculating \(P(z > z_{critical} - \Delta)\).
07

Evaluate Sample Size Adequacy

Compare the power calculated in Step 6 to a generally accepted standard such as 0.8 or 0.9. A power less than 0.8 suggests the sample size is not adequate to reliably detect a 12 psi difference, indicating the need for a larger sample size.

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Key Concepts

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

Confidence Interval
A confidence interval is a statistical tool used to estimate the range within which a population parameter lies, based on sample data. In this exercise, we're interested in the difference in mean breaking strengths between two types of plastic. The 95% confidence interval gives us a range that, if we conducted many samplings, we would expect the true mean difference to fall within 95% of the time. To compute this interval, we start from the observed difference in sample means and then add a margin of error. The margin of error is derived from the standard deviation and the critical value of the desired confidence level.

The formula used is: \[CI = (\bar{x}_1 - \bar{x}_2) \pm z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]This accounts for variability in the sample data due to the standard deviations and the size of the samples. A larger sample size decreases the margin of error, leading to a narrower confidence interval, giving us a more precise estimate.
Power of a Test
The power of a test is crucial in hypothesis testing as it tells us the probability of correctly rejecting the null hypothesis when it is false. In simple terms, it measures a test's ability to detect an actual effect when one exists. A higher power means a higher likelihood of spotting this effect, which in this case, is the difference in breaking strengths between the two plastics of at least 10 psi.

Power is influenced by several factors:
  • Sample Size: More data improves power.
  • Effect Size: Larger differences are easier to detect.
  • Significance Level \(\alpha\): Smaller values increase the threshold for significance but lower power.
To calculate the power, we use the non-centrality parameter, \(\Delta\), and compare it with our critical z-value. If the power is less than conventional thresholds like 0.8 or 0.9, the test might not reliably detect a true difference, as indicated in this exercise.
Sample Size Determination
In any statistical analysis, determining an adequate sample size is essential as it directly impacts the reliability and validity of the experiment's results. In our scenario, we're examining whether the sizes of 10 and 12 samples for plastic types 1 and 2 are sufficient to detect a meaningful difference in breaking strength, specifically a 12 psi difference.

Factors influencing sample size include:
  • The desired power of the test: Higher power requires larger samples.
  • The expected effect size: Smaller differences need larger samples to be significant.
  • The level of variability in the data: Greater variability requires more observations for accurate estimates.
If the calculated power from the existing sample sizes is inadequate (less than 0.8), it suggests gathering more data to increase the chances of detecting the difference reliably, thereby ensuring the conclusions drawn are robust and credible.

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