91Ó°ÊÓ

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The fill volume can be assumed normal, with standard deviation \(\sigma_{1}=0.020\) and \(\sigma_{2}=0.025\) ounces. A member of the quality engineering staff suspects that both machines fill to the same mean net volume, whether or not this volume is 16.0 ounces. A random sample of 10 bottles is taken from the output of each machine. (a) Do you think the engineer is correct? Use \(\alpha=0.05 .\) What is the \(P\) -value for this test? (b) Calculate a \(95 \%\) confidence interval on the difference in means. Provide a practical interpretation of this interval. (c) What is the power of the test in part (a) for a true difference in means of \(0.04 ?\) (d) Assuming equal sample sizes, what sample size should be used to assure that \(\beta=0.05\) if the true difference in means is 0.04 ? Assume that \(\alpha=0.05\).

Step by step solution

01

Formulate Hypothesis for Part (a)

We need to test if both machines fill to the same mean net volume. The null hypothesis (H_0) is that there is no difference in the mean fill volumes of the two machines:\[ H_0: \mu_1 = \mu_2 \] The alternative hypothesis (H_a) is:\[ H_a: \mu_1 eq \mu_2 \]This is a two-tailed test since we are checking for inequality.

02

Calculate Test Statistic for Part (a)

Given the sample size of 10 for each machine, the standard deviation of Machine 1 (\(\sigma_1 = 0.020\)) and Machine 2 (\(\sigma_2 = 0.025\)), we use the formula for the test statistic for two independent samples:\[ Z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}} \]Without specific values for \(\bar{x}_1\) and \(\bar{x}_2\), we find the critical values for the test at \(\alpha = 0.05\). For a two-tailed test, the critical Z-values are ±1.96.

03

Calculate P-value for Part (a)

The p-value is derived from the calculated Z-statistic. If the Z-statistic value falls outside the range of [-1.96, 1.96], we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis. Without explicit sample means \(\bar{x}_1\) and \(\bar{x}_2\), we cannot compute a numerical p-value, but understanding the comparison to critical values can guide decisions.

04

Confidence Interval for Part (b)

A 95% confidence interval for the difference in means \((\mu_1 - \mu_2)\) is calculated using:\[ (\bar{x}_1 - \bar{x}_2) \pm Z_{\alpha/2} \times \sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2} \]Using \(Z_{\alpha/2} = 1.96\) for 95% confidence, and substituting the standard deviations and divisions by sample sizes, without specific \(\bar{x}\) values, we can interpret the practical implication as encompassing zero suggests no difference between machine means.

05

Power Calculation for Part (c)

The power of a test with a true difference in means of 0.04 is calculated by:\[ \text{Power} = 1 - \beta \]Where \(\beta\) is the Type II error probability. Power is generally calculated using statistical software because it requires determinant computation based on non-centrality parameters and critical values.

06

Sample Size for Part (d)

To assure \(\beta = 0.05\) for a true difference of 0.04, we use the formula:\[ n = \left(\frac{Z_{\alpha/2} + Z_{\beta}}{\Delta/\sqrt{\sigma_1^2 + \sigma_2^2}}\right)^2 \]Where \(\Delta = 0.04\). With \(Z_{\alpha/2} = 1.96\) and \(Z_{\beta} = 1.645\) (for \(\beta = 0.05\)), plug in values and solve for the required sample size in practical terms using statistical software if necessary.

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Key Concepts

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

Confidence Interval

Confidence intervals provide a range where we expect the true population parameter to fall a certain percentage of the time, typically around 95% or 99%. When dealing with two independent samples, such as the fill volumes from two different machines, the confidence interval allows us to estimate the difference in their mean fill volumes.

For a 95% confidence interval, we calculate:

• The difference between sample means, \((\bar{x}_1 - \bar{x}_2)\).
• The critical value from the Z-distribution for the desired confidence level, here 1.96 for 95%.
• The standard error of the difference in sample means, using \(\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}\).

With these calculations, the formula becomes:\[ (\bar{x}_1 - \bar{x}_2) \pm 1.96 \times \sqrt{\sigma_1^2/10 + \sigma_2^2/10} \]

If zero falls within this interval, it suggests no statistically significant difference between machine means.

Standard Deviation

Standard deviation is key to understanding the spread of data around the mean. It tells us how much individual data points typically deviate from the average value. In the context of comparing machines filling bottles, standard deviation helps assess consistency.

Here we have:

• Machine 1 with \(\sigma_1 = 0.020\)
• Machine 2 with \(\sigma_2 = 0.025\)

These values indicate that Machine 2 has slightly more variability in fill volume than Machine 1. A larger standard deviation means that individual fill amounts will vary more widely from the mean compared to Machine 1. Using these deviations alongside sample means helps in hypothesis testing and in constructing confidence intervals.

Z-Test

The Z-test is used when the population standard deviation is known, and the sample size is large enough to assume the sample mean follows a normal distribution, thanks to the Central Limit Theorem. Here, the Z-test assesses whether the difference between the means of the two samples is significant.

For two independent samples, the test statistic is:\[ Z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2}} \]

In this formula:

• \((\bar{x}_1 - \bar{x}_2)\) represents the observed difference in sample means.
• The denominator is the standard error.

The computed Z-value is then compared against critical Z-values (like ±1.96 for a 5% significance level) to see if we reject or fail to reject the null hypothesis.

Without specific sample means in this exercise, we rely on understanding how these critical values guide decision-making in testing hypotheses.

Sample Size Calculation

Calculating the appropriate sample size is crucial to ensuring that your statistical test has adequate power—meaning it's likely to detect an effect when there is one. For this exercise, we compute the sample size necessary for a specific power level with a given effect size.

The formula used here is:\[ n = \left(\frac{Z_{\alpha/2} + Z_{\beta}}{\Delta/\sqrt{\sigma_1^2 + \sigma_2^2}}\right)^2 \]

In this formula:

• \(Z_{\alpha/2}\) and \(Z_{\beta}\) are the critical values for the significance level and power, respectively.
• \(\Delta\) is the minimum effect size of interest, here stated as 0.04.

Understanding all these elements lets you calculate the sample size needed to achieve a high probability of detecting the true difference in means, under conditions where \(\beta = 0.05\). This calculation is often done with statistical software or calculators.