/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A spectrophotometer used for mea... [FREE SOLUTION] | 91Ó°ÊÓ

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A spectrophotometer used for measuring \(\mathrm{CO}\) concentration [ppm (parts per million) by volume] is checked for accuracy by taking readings on a manufactured gas (called span gas) in which the \(\mathrm{CO}\) concentration is very precisely controlled at \(70 \mathrm{ppm}\). If the readings suggest that the spectrophotometer is not working properly, it will have to be recalibrated. Assume that if it is properly calibrated, measured concentration for span gas samples is normally distributed. On the basis of the six readings- \(85,77,82,68,72\), and 69-is recalibration necessary? Carry out a test of the relevant hypotheses using the \(P\)-value approach with \(\alpha=.05\).

Short Answer

Expert verified
Recalibration is not necessary.

Step by step solution

01

Identify the Hypotheses

First, we need to set up our null and alternative hypotheses for the hypothesis test. The null hypothesis \( H_0 \) states that the spectrophotometer is properly calibrated, i.e., the true mean of the readings is equal to the known concentration of 70 ppm. The alternative hypothesis \( H_a \) states that the spectrophotometer is not properly calibrated, i.e., the true mean of the readings is not equal to 70 ppm ([two-tailed test](https://en.wikipedia.org/wiki/Statistical_hypothesis_testing)). Thus, \( H_0: \mu = 70 \) and \( H_a: \mu eq 70 \).
02

Calculate Sample Statistics

Next, calculate the sample mean \( \bar{x} \) and the sample standard deviation \( s \). Given the six readings \( 85, 77, 82, 68, 72, \) and \( 69 \), we find:- The sample mean: \( \bar{x} = \frac{85 + 77 + 82 + 68 + 72 + 69}{6} = 75.5 \).- The sample variance \( s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 \), calculate this, then take the square root for the standard deviation \( s \).
03

Compute the Test Statistic

Now compute the test statistic \( t \) using the formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]where \( \mu_0 \) is 70 ppm, \( \bar{x} \) is 75.5 ppm, \( s \) is the sample standard deviation, and \( n \) is the number of samples (6).
04

Find the P-value

Determine the \( P \)-value corresponding to the computed test statistic. Since this is a two-tailed test, find the probability that the test statistic is as extreme as, or more extreme than, the observed value under \( H_0 \). Use the \( t \)-distribution with \( n-1 = 5 \) degrees of freedom to find this probability.
05

Make a Decision

Compare the computed \( P \)-value to the significance level \( \alpha = 0.05 \). If the \( P \)-value is less than or equal to 0.05, reject the null hypothesis. If the \( P \)-value is greater than 0.05, do not reject the null hypothesis. This decision tells us whether recalibration is necessary.
06

Conclusion

Based on the computed \( P \)-value and comparison to \( \alpha \), we found that the \( P \)-value was greater than 0.05, so we do not reject the null hypothesis. Therefore, recalibration of the spectrophotometer is not necessary.

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Key Concepts

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

Two-Tailed Test
A two-tailed test is a type of hypothesis test used in statistics to determine if a sample is significantly different from the null hypothesis. It checks whether the true mean or proportion differs from the hypothesized value in either direction. In our case, the null hypothesis posits that the mean concentration given by the spectrophotometer is 70 ppm. The alternative hypothesis suggests that it is "not equal to" 70 ppm, meaning it could be either above or below this value.
A two-tailed test is appropriate when we are concerned about deviations in both directions, which is why it was used for determining if the spectrophotometer needed recalibration.
The significance level, often denoted by \( \alpha \) and usually set at 0.05, dictates the probability threshold for rejecting the null hypothesis. In a two-tailed test, this \( \alpha \) level is split between the two tails of the distribution, which makes the decision more comprehensive.
P-Value Approach
The P-Value approach allows us to make decisions about the null hypothesis by evaluating the probability of observing a test statistic as extreme as the one obtained, under the assumption that the null hypothesis is true.
In our spectrophotometer example, once we calculate the test statistic using the sample data, we reference this value in a \( t \)-distribution table. We then determine the P-Value, which indicates the likelihood of obtaining a sample mean of 75.5 ppm, or further away from 70 ppm, if the true mean is indeed 70 ppm.
If this P-value is less than or equal to our chosen significance level \( \alpha = 0.05 \), we reject the null hypothesis, concluding that recalibration is necessary. Conversely, a P-value greater than 0.05 implies that there is not enough statistical evidence to claim that the spectrophotometer requires recalibration.
Null and Alternative Hypotheses
In any hypothesis test, formulating the correct null and alternative hypotheses is crucial. The null hypothesis \( H_0 \) generally presents a statement of no change or effect and is believed to be true until statistical evidence indicates otherwise. Here, \( H_0: \mu = 70 \) asserts that the spectrophotometer's readings are accurate and require no recalibration.
The alternative hypothesis \( H_a \), on the other hand, challenges the status quo presented by the null hypothesis. It posits \( H_a: \mu eq 70 \), indicating a suspicion that the spectrophotometer might not accurately measure 70 ppm, thus needing recalibration.
Choosing the correct hypotheses ensures the test examines the relevant questions and makes informed decisions about device accuracy.
t-Distribution
The \( t \)-distribution is a probability distribution that is used when the sample size is small, typically less than 30, and when the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, which makes it more accommodating for small sample sizes by accounting for variability.
In our example, because we only have six measurements from the spectrophotometer, the \( t \)-distribution is appropriate. We use this distribution to find critical values and correspond the calculated test statistic to determine the P-value.
The degrees of freedom, calculated as the sample size minus one (\( n-1 \)), help tailor the \( t \)-distribution to the specific sample. In our case with 6 samples, the distribution uses 5 degrees of freedom.
Statistical Decision Making
Statistical decision making involves using data to make informed conclusions about hypotheses. In hypothesis testing, statistical decision making helps decide whether to reject or not reject the null hypothesis. The decision is based on comparing the P-value to the predefined significance level \( \alpha \).
If the P-value is less or equal to \( \alpha \), it indicates enough evidence against \( H_0 \), leading to its rejection. If the P-value is greater, the null hypothesis is not rejected, suggesting there is not enough evidence to support \( H_a \).
In the spectrophotometer scenario, statistical decision making concluded that the device did not need recalibration as the P-value exceeded the significance level, indicating no significant deviation from the expected 70 ppm readings.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from an exponential distribution with parameter \(\lambda\). Then it can be shown that \(2 \lambda \Sigma X_{i}\) has a chi-squared distribution with \(v=2 n\) (by first showing that \(2 \lambda X_{i}\) has a chi-squared distribution with \(v=2\) ). a. Use this fact to obtain a test statistic and rejection region that together specify a level \(\alpha\) test for \(H_{0}: \mu=\mu_{0}\) versus each of the three commonly encountered alternatives. [Hint: \(E\left(X_{i}\right)=\) \(\mu=1 / \lambda\), so \(\mu=\mu_{0}\) is equivalent to \(\lambda=1 / \mu_{0}\).] b. Suppose that ten identical components, each having exponentially distributed time until failure, are tested. The resulting failure times are \(\begin{array}{llllllllll}95 & 16 & 11 & 3 & 42 & 71 & 225 & 64 & 87 & 123\end{array}\) Use the test procedure of part (a) to decide whether the data strongly suggests that the true average lifetime is less than the previously claimed value of 75 .

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