Two machines are used for filling plastic bottles with a net...

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 $\alpha=0.05$.

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 $\alpha=0.05$.

Key Concepts

Type I and Type II Errors

Type I error (?) is the probability of incorrectly rejecting the true null hypothesis, while Type II error (?) is the probability of failing to reject a false null hypothesis. These concepts are critical in hypothesis testing as they relate to the risks of false positives and false negatives, respectively. Understanding these errors is essential for designing tests, interpreting results, and determining sample sizes to achieve acceptable levels of error.

Sample Size Determination

This concept involves calculating the minimum number of subjects or observations needed to achieve a desired statistical power (often 1-?, such as 0.95) for detecting a specified effect size at a given significance level (?). In the context of comparing two means, the sample size determination accounts for the variability in both groups and the required precision to ensure that the study can reliably detect a true difference if it exists.

Statistical Power

Statistical power is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true, i.e., detecting a true effect when it exists. Power depends on several factors including the sample size, the magnitude of the effect (difference in means), the significance level (?), and the variability in the data. Assessing power is important to understand the sensitivity of the test and to plan for appropriate sample sizes.

P-Value

The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is true. It quantifies the evidence against the null hypothesis; a smaller p-value indicates stronger evidence to reject H0. In the context of comparing means, the p-value is derived from the sampling distribution of the test statistic (often using a z-test or t-test depending on the situation).

Hypothesis Testing for Difference in Means

This concept involves comparing the means from two independent populations to determine if there is a statistically significant difference between them. It typically requires setting up null and alternative hypotheses (e.g., H0: the means are equal, Ha: the means are not equal) and calculating a test statistic that reflects how far apart the sample means are relative to the standard error of the difference, assuming the null hypothesis is true.

Confidence Interval for Difference in Means

A confidence interval provides a range of plausible values for the true difference between population means. Constructed by taking the estimated difference and adding and subtracting a margin of error (which is determined by the standard error and a critical value from the appropriate distribution), it offers an interval estimate that, with a specified level of confidence (e.g., 95%), is likely to contain the true difference.

Transcript

00:01 Hi guys in this problem we apply the seven step procedures to this problem as follows so the first step is to determine the parameter of interest the parameter of interest here is the quantity of interest is the difference in mean volume so it's mu 1 minus mu 2 and we have delta node equal 0 so the null hypothesis such that mu 1 minus mu note equals 0 okay and the alternative hypothesis h1 okay it's mu 1 minus mu 2 does not equal 0 okay so we want to reject the anal hypothesis if there is a difference in mean volume so as a test statistics z note equals x1 par minus x2 par time minus delta node over the square root of sigma 1 squared over n1 plus sigma 2 squared over n2 okay so we have sigma 1 squared over equals 0 .02 squared okay sigma 2 squared is just 0 .022 squared okay square okay so and we have n1 plus equals n2 equals 10 okay so by using the p value approach we reject the null hypothesis if the p value is less than 0 .05 okay so computations from the same build data let's compute x1 dash it's a summation of x1 1 over n1 okay so it's 16 .03 plus 16 .04 and so on over 10 okay so x1 -equals 16 .15 okay so x1 -1 equals 16 .1015 okay so x2 dash equals summation over x2 over n so it's 16 .02 plus 15 .97 and so on, okay, over 10.

02:53 So x2 equals 16 .005.

02:58 So if we re -substitute in the z -note formula, we get z -note equal 0 .99.

03:07 Then to conclude for these two tail tests, the p value is 2 times 1 minus 5 of 0 .99.

03:23 Okay, so this is 0 .3222.

03:30 Okay, since the p value is much more the 0 .05, we fail to reject the null hypothesis.

03:37 Therefore, the engineer is correct.

03:39 The pulse machines fell the same mean net volume, regardless of whether the volume is 16 pounds.

03:46 Okay.

03:49 Then for part p, the data obtained as follows.

03:53 We have x1 -dash, x2, sigma -1 and sigma -2, and n -1 and n2.

03:59 So the critical value of z at 95 % confidence level is z of alpha over 2.

04:06 So it's 1 .96.

04:11 Okay, so let's calculate x1 par minus x2 power, plus or minus z of alpha over 2, 4 times sigma 1 squared over n1 plus sigma 2 squared over n.

04:39 Okay, so this is 0 .01 plus or minus o .1 plus or minus 0 .02...

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