91Ó°ÊÓ

At Farmer's Dairy, a machine is set to fill 32 -ounce milk cartons. However, this machine does not put exactly 32 ounces of milk into each carton; the amount varies slightly from carton to carton but has a normal distribution. It is known that when the machine is working properly, the mean net weight of these cartons is 32 ounces. The standard deviation of the milk in all such cartons is always equal to \(.15\) ounce. The quality control inspector at this company takes a sample of 25 such cartons every week, cal culates the mean net weight of these cartons, and tests the null hypothesis, \(\mu=32\) ounces, against the alternative hypothesis, \(\mu \neq 32\) ounces. If the null hypothesis is rejected, the machine is stopped and ad. justed. A recent sample of 25 such cartons produced a mean net weight of \(31.93\) oun Calculate the \(p\) -value for this test of hypothesis. Based on this \(p\) -value, will the quality control. inspector decide to stop the machine and readjust it if she chooses the maximum probability of Type I error to be \(.01\) ? What if the maximum probability of a Type I error is b. Test the hypothesis of part a using the critical-value approach and \(\alpha=.01 .\) Does the machine need to be adjusted? What if \(\alpha=05\) ?

Step by step solution

01

Calculate Test Statistic

We first calculate the test statistic (z-score) using the given sample mean (31.93 ounces), population mean (32 ounces), standard deviation (0.15 ounces), and sample size (25). The formula to calculate z-score is \( z = \frac{{\(\bar{x}\) - \(\mu\)}}{{\(\sigma/\sqrt{n}\)}} \)Using the given values, we get \( z = \frac{{31.93 - 32}}{{0.15/\sqrt{25}}}= -2.33 \)

02

Calculating the p-value

We look up the z-score (-2.33) in the z-table (or use a statistical software) to find the probability. Because this is a two-tailed test, we calculate the two-tail probability (p-value) by multiplying the one-tail probability by 2. The corresponding one-tail probability for -2.33 is approximately 0.0099. Thus, the two-tail p-value is 0.0099 * 2 = 0.0198.

03

Decision Based on p-value

Now, we need to compare the p-value with the maximum Type I error rates. If the p-value is less than the error rate, we reject the null hypothesis and decide to adjust the machine. Here, for the error rate of 0.01, the p-value (0.0198) is greater, so we do not reject the null hypothesis and don't adjust the machine. However, for a Type I error rate more than 0.0198, we would reject the null hypothesis and decide to adjust the machine.

04

Test Using Critical-value Approach

We find critical z-scores for a two-tailed test at alpha level 0.01. The critical z-scores are -2.58 and +2.58. Since -2.33 lies between these two values, we do not reject the null hypothesis at alpha =0.01. For alpha = 0.05, the critical z-scores are -1.96 and +1.96. Here, -2.33 falls outside this range, hence we reject the null hypothesis and decide to adjust the machine.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution

The normal distribution, often called the bell curve, represents how data is spread out, with most values clustering around the mean. It is symmetrical and describes many natural phenomena. For Farmer's Dairy, the milk carton weights follow this kind of distribution, centered around a mean of 32 ounces. This means most cartons are filled close to 32 ounces, but there is variation modeled by the standard deviation of 0.15 ounces.
Understanding the normal distribution helps us predict how likely certain outcomes are based on the mean and standard deviation. This is crucial for quality control, ensuring that the machine's output remains consistent over time.

Type I Error

A Type I error occurs when you reject the null hypothesis when it is actually true. This is like thinking something is wrong with the machine when it’s working fine. You can think of it as a false alarm.
The probability of making a Type I error is denoted by alpha \( \alpha \). Here, the quality control inspector sets this probability to 0.01 or 0.05, depending on how cautious they want to be. A lower alpha means you want to be more certain before deciding to adjust the machine, minimizing unnecessary stoppages.

Z-score

The z-score is a measure that tells us how many standard deviations an element is from the mean. In this case, we use the formula \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \) to calculate the z-score. Here, \( \bar{x} \) is the sample mean (31.93 ounces), \( \mu \) is the population mean (32 ounces), \( \sigma \) is the standard deviation (0.15 ounces), and \( n \) is the sample size (25).
The resulting z-score of -2.33 indicates how the sample mean deviates from the population mean. This helps in making decisions about the null hypothesis, with more extreme z-values suggesting that something unusual is happening with the machine.

Critical Value Approach

The critical value approach involves comparing the calculated z-score to critical values that define the boundaries of the acceptance region under the null hypothesis. For a two-tailed test:

• At \( \alpha = 0.01 \), critical values are -2.58 and +2.58, meaning we reject the null if the z-score is beyond these values.
• At \( \alpha = 0.05 \), these critical values are -1.96 and +1.96.

If the z-score falls between the critical values, we do not reject the null hypothesis. For Farmer's Dairy, a z-score of -2.33 falls within the critical range for \( \alpha = 0.01 \) but not for \( \alpha = 0.05 \). This means the decision to adjust the machine depends on the chosen significance level, illustrating how different thresholds for Type I error impact decision-making.