91Ó°ÊÓ

Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let \(p\) denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than \(20 \%\) of all specimens yield before the theoretical point, the production process will have to be modified. a. If 15 of 60 specimens yield before the theoretical point, what is the \(P\)-value when the appropriate test is used, and what would you advise the company to do? b. If the true percentage of "early yields" is actually \(50 \%\) (so that the theoretical point is the median of the yield distribution) and a level \(.01\) test is used, what is the probability that the company concludes a modification of the process is necessary?

Step by step solution

01

Calculate Sample Proportion

The sample proportion \( \hat{p} \) is calculated by dividing the number of successes (specimens yielding early) by the total number of specimens. For this problem, \( \hat{p} = \frac{15}{60} = 0.25 \).

02

State Null and Alternative Hypotheses

The null hypothesis \( H_0 \) is that \( p \leq 0.2 \), and the alternative hypothesis \( H_a \) is that \( p > 0.2 \). We are testing if more than 20% yield early.

03

Determine Test Statistic

Assuming the sampling distribution of \( \hat{p} \) is approximately normal, calculate the test statistic using the formula \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \), where \( p_0 = 0.2 \). Thus, \[ z = \frac{0.25 - 0.2}{\sqrt{\frac{0.2 \times 0.8}{60}}} \approx 1.095. \]

04

Find the P-Value

Using the standard normal distribution table, the \( P \)-value for \( z = 1.095 \) is approximately 0.1368. This represents the probability of observing a sample proportion at least as extreme as \( 0.25 \), assuming the true proportion is 0.2.

05

Decision Based on P-Value (Part a)

Since the \( P \)-value of 0.1368 is greater than any typical significance level like 0.05 or 0.01, we fail to reject \( H_0 \). The company does not need to modify the production process based on this sample.

06

Probability of Type I Error (Part b)

Given that \( p = 0.5 \), under the null hypothesis, we need to calculate the probability that \( \hat{p} > 0.2 \). Using a normal approximation, the standard error is \( \sqrt{\frac{0.5 \times 0.5}{60}} \approx 0.0645 \), and find \( z \) for \( \hat{p} = 0.2 \), which is \( z = \frac{0.2 - 0.5}{0.0645} = -4.65 \). Thus, \( P(\hat{p} > 0.2) = P(z > -4.65) \) which is essentially 1, implying the process modification will often be deemed necessary even if \( p = 0.5 \) due to a typographical error.

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Key Concepts

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

Sample Proportion

When trying to understand a large group by examining a smaller subset, we calculate something called the sample proportion, represented by \( \hat{p} \). This is simply the fraction or percentage of interest that appears in your sample. For example, in a study sample of 60 steel specimens, if 15 yield early, the sample proportion is found as \( \hat{p} = \frac{15}{60} \) equaling 0.25 or 25%.

To calculate this, divide the number of "successes" (in this case, early yield samples) by the total number of samples. The sample proportion gives us an estimate of the true proportion for the entire population, which would be all possible specimens in the factory. This estimation helps make decisions about whether our production process needs adjusting.

P-Value

The P-value in statistical hypothesis testing is a measure that helps determine the strength of your results. Essentially, it tells you how likely it is to observe the outcome of your sample, assuming that the null hypothesis is true.

In our example, the null hypothesis suggests that only 20% of specimens should yield early. After calculating the P-value, which was approximately 0.1368 in this instance, it reflects the probability of seeing a sample proportion as extreme as 0.25 due to random chance alone. If the P-value is small (usually less than 0.05), it suggests that such an extreme observation is unlikely when the null hypothesis is true, leading us to question the null hypothesis.

Null Hypothesis

The null hypothesis, often represented as \( H_0 \), is a statement used in hypothesis testing that assumes there is no effect or difference. It provides a baseline for statistical comparison. In this exercise, the null hypothesis posits that the true sample proportion \( p \) is less than or equal to 0.20.

The alternative hypothesis, represented as \( H_a \), notes that \( p \) is more than 0.20, indicating a need for a process modification. Hypothesis testing aims to determine whether there is enough statistical evidence to reject the null hypothesis. Remember, failing to reject \( H_0 \) does not prove it true, it simply means there's not enough evidence to support \( H_a \).

Type I Error

A Type I error occurs in hypothesis testing when the null hypothesis is incorrectly rejected when it is actually true. This is also known as a "false positive."

In context, if the company decides to modify the production process believing that more than 20% of specimens yield early, but in reality, it's not more than 20%, a Type I error has occurred. The significance level \( \alpha \), such as 0.01 or 0.05, is the probability of making this error. It's important to carefully choose this threshold to balance the risk of rejecting a true null hypothesis. The smaller \( \alpha \) is, the less likely a Type I error, but this may increase the risk of a Type II error (failing to reject a false null hypothesis).

Understanding these concepts allows you to make informed decisions based on statistical data—whether you're testing steel or any other hypothesis.