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In the past, a chemical plant has produced an average of 1100 pounds of chemical per day. The records for the past year, based on 260 operating days, show the following: $$\bar{y}=1060 \text { pounds/day, } \quad s=340 \text { pounds/day }$$ We wish to test whether the average daily production has dropped significantly over the past year. a. Give the appropriate null and alternative hypotheses. b. If \(Z\) is used as a test statistic, determine the rejection region corresponding to a level of significance of \(\alpha=.05\) c. Do the data provide sufficient evidence to indicate a drop in average daily production?

Short Answer

Expert verified
Reject the null hypothesis; the average daily production has significantly dropped.

Step by step solution

01

Define the Hypotheses

To test whether there has been a significant drop in average daily production, we define the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). \(H_0\): The average production per day is 1100 pounds (\(\mu = 1100\)). \(H_a\): The average production per day is less than 1100 pounds (\(\mu < 1100\)).
02

Identify the Test Statistic and Level of Significance

We use the \(Z\) statistic for this hypothesis test since the sample size is large (\(n = 260\)). The test statistic is calculated using \(Z = \frac{\bar{y} - \mu}{s/\sqrt{n}}\). The level of significance \(\alpha = 0.05\).
03

Determine the Rejection Region

For a one-tailed test at \(\alpha = 0.05\), the rejection region is the value of \(Z\) less than the critical value (\(Z_{\alpha}\)). From the standard normal distribution table, \(Z_{\alpha} = -1.645\). Hence, the rejection region is \(Z < -1.645\).
04

Compute the Test Statistic

Calculate the \(Z\) value using the formula: \(Z = \frac{1060 - 1100}{340/\sqrt{260}}\). First, compute the standard error: \(SE = \frac{340}{\sqrt{260}}\approx 21.12\). Then, compute \(Z = \frac{1060 - 1100}{21.12} \approx -1.89\).
05

Make a Decision

Compare the calculated \(Z\) value with the critical \(Z\) value. Since \(Z = -1.89\) is less than \(-1.645\), we reject the null hypothesis. This indicates there is sufficient evidence to conclude that the average daily production has dropped significantly.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is the foundation of hypothesis testing. It is a statement made to express that there is no effect or no difference in a particular situation. In our given problem, the null hypothesis asserts that the average production per day in the chemical plant has remained the same, specifically at 1100 pounds. Mathematically expressed, it is \(H_0: \mu = 1100\).

The significance of the null hypothesis is its role as the benchmark against which new claims, like changes in production rates, are compared. When conducting a hypothesis test, we usually assume the null hypothesis is true until evidence suggests otherwise.

In practice, the null hypothesis is vital because it provides a baseline expectation that is used to determine the significance of the test's observed data. Rejection or non-rejection of \(H_0\) can greatly affect decisions or inferences drawn from the data.
Alternative Hypothesis
The alternative hypothesis, represented as \(H_a\), is a statement that contradicts the null hypothesis. It suggests that there is a change, a difference, or a new effect that requires attention. In our context, the alternative hypothesis is that the average daily production is less than 1100 pounds. This is expressed as \(H_a: \mu < 1100\).

Choosing the right alternative hypothesis is crucial because it guides the direction of the test. Since we are interested in finding out if there is a significant drop in production, a one-tailed hypothesis is appropriate. The choice between one-tailed and two-tailed tests depends on the problem specifics and the research question being addressed.

The alternative hypothesis helps drive the entire testing process and offers a contrast to the \(H_0\). If the test results show strong evidence against \(H_0\), then \(H_a\) becomes more plausible based on the chosen level of significance.
Z Statistic
The \(Z\) statistic is a part of the standard normal distribution used to determine the position of a sample mean relative to the population mean in hypothesis testing. It is particularly useful when the sample size is large (usually \(n > 30\)). In our scenario with the chemical plant, the \(Z\) statistic is calculated using the formula: \[ Z = \frac{\bar{y} - \mu}{s/\sqrt{n}} \]where \(\bar{y}\) is the sample mean, \(\mu\) is the population mean according to \(H_0\), \(s\) is the standard deviation of the sample, and \(n\) is the sample size.

The \(Z\) statistic is crucial because it allows us to compare the observed test result to what we would expect if the null hypothesis were true. In simpler terms, it measures how many standard errors the sample mean is away from the population mean.

After calculating the \(Z\) value, we compare it to critical values from the standard normal distribution table to decide whether to reject \(H_0\). An extreme \(Z\) statistic indicates that the null hypothesis might not hold.
Level of Significance
The level of significance, denoted as \(\alpha\), is a threshold set by the researcher that determines when we should reject the null hypothesis. It is typically set at 0.05, 0.01, or 0.10. In this exercise, \(\alpha = 0.05\) is chosen, which signifies a 5% risk of rejecting the null hypothesis when it is true.

Setting the level of significance is important because it reflects the researcher's tolerance for error. A lower \(\alpha\) means the researcher demands stronger evidence to reject \(H_0\).

The level of significance also helps in defining the rejection region. In a one-tailed test like the one we're dealing with, \(\alpha\) determines the critical value. If our test statistic (like \(Z\)) falls into this rejection region, it means the data provide enough evidence to reject the null hypothesis.

Choosing \(\alpha\) thus directly affects how we interpret results and the conclusions we draw regarding the hypotheses being tested.

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Most popular questions from this chapter

The stability of measurements of the characteristics of a manufactured product is important in maintaining product quality. In fact, it is sometimes better to obtain small variation in the measured value of some important characteristic of a product and have the process mean slightly off target than to get wide variation with a mean value that perfectly fits requirements. The latter situation may produce a higher percentage of defective product than the former. A manufacturer of light bulbs suspected that one of his production lines was producing bulbs with a high variation in length of life. To test this theory, he compared the lengths of life of \(n=50\) bulbs randomly sampled from the suspect line and \(n=50\) from a line that seemed to be in control. The sample means and variances for the two samples were as shown in the following table. $$\begin{array}{ll} \hline \text { Suspect Line } & \text { Line in Control } \\ \hline \bar{y}_{1}=1,520 & \bar{y}_{2}=1,476 \\ s_{1}^{2}=92,000 & s_{2}^{2}=37,000 \\ \hline \end{array}$$ a. Do the data provide sufficient evidence to indicate that bulbs produced by the suspect line possess a larger variance in length of life than those produced by the line that is assumed to be in control? Use \(\alpha=.05\) b. Find the approximate observed significance level for the test and interpret its value.

A political researcher believes that the fraction \(p_{1}\) of Republicans strongly in favor of the death penalty is greater than the fraction \(p_{2}\) of Democrats strongly in favor of the death penalty. He acquired independent random samples of 200 Republicans and 200 Democrats and found 46 Republicans and 34 Democrats strongly favoring the death penalty. Does this evidence provide statistical support for the researcher's belief? Use \(\alpha=.05\)

The state of California is working very hard to ensure that all elementary age students whose native language is not English become proficient in English by the sixth grade. Their progress is monitored each year using the California English Language Development test. The results for two school districts in southern California for the 2003 school year are given in the accompanying table. \(^{\star}\) Do the data indicate a significant difference in the 2003 proportions of students who are fluent in English for the two districts? Use \(\alpha=.01\) $$\begin{array}{lcc} \hline \text { District } & \text { Riverside } & \text { Palm Springs } \\ \hline \text { Number of students tested } & 6124 & 5512 \\ \text { Percentage fluent } & 40 & 37 \\ \hline \end{array}$$

True or False. a. If the \(p\) -value for a test is .036 , the null hypothesis can be rejected at the \(\alpha=.05\) level of significance. b. In a formal test of hypothesis, \(\alpha\) is the probability that the null hypothesis is incorrect. c. If the \(p\) -value is very small for a test to compare two population means, the difference between the means must be large. d. Power \(\left(\theta^{*}\right)\) is the probability that the null hypothesis is rejected when \(\theta=\theta^{*}\). e. Power( \((\theta)\) is always computed by assuming that the null hypothesis is true. f. If \(.01 < p\) -value \( < .025\), the null hypothesis can always be rejected at the \(\alpha=.02\) level of significance. g. Suppose that a test is a uniformly most powerful \(\alpha\) -level test regarding the value of a parameter \(\theta .\) If \(\theta_{a}\) is a value in the alternative hypothesis, \(\beta\left(\theta_{a}\right)\) might be smaller for some other \(\alpha\) -level test. h. When developing a likelihood ratio test, it is possible that \(L\left(\widehat{\Omega}_{0}\right) > L(\widehat{\Omega})\) i. \(-2 \ln (\lambda)\) is always positive.

Why is the \(Z\) test usually inappropriate as a test procedure when the sample size is small?

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