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A sample of 50 lenses used in eyeglasses yields a sample mean thickness of \(3.05 \mathrm{~mm}\) and a sample standard deviation of \(.34 \mathrm{~mm}\). The desired true average thickness of such lenses is \(3.20 \mathrm{~mm}\). Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using \(\alpha=.05\).

Short Answer

Expert verified
The data strongly suggests the average thickness is different from 3.20 mm.

Step by step solution

01

Define the Hypotheses

The first step in hypothesis testing is to define the null and alternative hypotheses. Here, the null hypothesis \(H_0\) states that the true average thickness \(\mu\) is equal to the desired thickness, \(\mu = 3.20\,\text{mm}\). The alternative hypothesis \(H_a\) states that the true average thickness is not equal to the desired thickness, \(\mu eq 3.20\,\text{mm}\). This is a two-tailed test.
02

Determine the Test Statistic

We use the t-test for the sample mean because the population standard deviation is unknown. The test statistic is calculated using the formula:\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x} = 3.05\,\text{mm}\) is the sample mean, \(\mu = 3.20\,\text{mm}\) is the population mean, \(s = 0.34\,\text{mm}\) is the sample standard deviation, and \(n = 50\) is the sample size.
03

Calculate the Test Statistic

Substituting the known values in, the test statistic \( t \) can be calculated as follows:\[ t = \frac{3.05 - 3.20}{0.34 / \sqrt{50}} = \frac{-0.15}{0.048} \approx -3.125 \]
04

Determine the Critical t-value

For a two-tailed test at a significance level \( \alpha = 0.05 \) with \( n - 1 = 49 \) degrees of freedom, we consult a t-distribution table. The critical t-value is roughly \( \pm 2.0096 \).
05

Make a Decision

Compare the calculated t-value \( -3.125 \) with the critical values \( \pm 2.0096 \). Since \( -3.125 \) falls outside the range \(-2.0096\) to \(2.0096\), we reject the null hypothesis \(H_0\).
06

Conclusion

By rejecting the null hypothesis, we conclude that the data strongly suggests that the true average thickness of the lenses differs from the desired thickness of \(3.20\,\text{mm}\).

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Key Concepts

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

Two-Tailed Test
When conducting a hypothesis test, a two-tailed test is applicable when you want to determine if there is a significant difference in either direction from a specified value. Here, the primary question is whether the true mean thickness of the lenses is different from the desired thickness of 3.20 mm, meaning it could be either greater or smaller.
Thus, we are interested in deviations on both sides, hence the "two-tailed" terminology.
This approach is comprehensive because it considers the possibility of an average that is higher or lower than expected, allowing for more flexibility in testing hypotheses.
t-Distribution
The t-distribution is used when the sample size is small, and the population standard deviation is unknown, which is the case in many real-world situations. For our problem, the sample size is 50, and the population standard deviation isn't known, so the t-distribution is appropriate.
The shape of the t-distribution is similar to the normal distribution but has heavier tails, allowing for greater variability with smaller samples.
As the sample size increases, the t-distribution approaches the normal distribution because estimates become more precise with larger datasets. The degrees of freedom, calculated as the sample size minus one, also affect the shape of the t-distribution, and in this example, there are 49 degrees of freedom.
Significance Level
In hypothesis testing, the significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is true. It is a predetermined threshold for significance.
For this exercise, a significance level of 0.05 is chosen, which is common in many fields of research.
This level represents a 5% risk of concluding that a difference exists when there is none.
Thus, it sets a high standard for evidence before concluding that the true mean thickness of the lenses differs from the anticipated 3.20 mm.
Sample Mean
The sample mean \( \bar{x} \) is a key measure of central tendency that provides an estimate of the population mean. In this scenario, the sample mean is 3.05 mm.
This value was calculated from the 50 lenses sampled and forms the basis of comparison against the desired mean of 3.20 mm.
The difference between the sample mean and the specified population mean helps us compute the test statistic and decide the outcome of our hypothesis test.
A sample mean that differs significantly from the population mean suggests that the null hypothesis may not hold.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. Here, the sample standard deviation is 0.34 mm.
This value indicates how much the individual lens thickness measurements deviate from the sample mean of 3.05 mm.
A small standard deviation implies that the sample data points are close to the mean, indicating lesser variability.
In hypothesis testing, the standard deviation is vital for calculating the test statistic, thus influencing the conclusions about the population mean.

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Most popular questions from this chapter

Let the test statistic \(Z\) have a standard normal distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}\), rejection region \(z \geq 1.88\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}\), rejection region \(z \leq-2.75\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), rejection region \(z \geq 2.88\) or \(z \leq\) \(-2.88\)

The recommended daily dietary allowance for zinc among males older than age 50 years is \(15 \mathrm{mg} /\) day. The article "Nutrient Intakes and Dietary Pattems of Older Americans: A National Study" (J. Gerontol., 1992: M145-150) reports the following summary data on intake for a sample of males age 65-74 years: \(n=115, \bar{x}=11.3\), and \(s=6.43\). Does this data indicate that average daily zinc intake in the population of all males age 65-74 falls below the recommended allowance?

Suppose that each of \(n\) randomly selected individuals is classified according to his/her genotype with respect to a particular genetic characteristic and that the three possible genotypes are \(\mathrm{AA}, \mathrm{Aa}\), and aa with long-run proportions (probabilities) \(\theta^{2}\), \(2 \theta(1-\theta)\), and \((1-\theta)^{2}\), respectively \((0<\theta<1)\). It is then straightforward to show that the likelihood is $$ \theta^{2 x_{1}} \cdot[2 \theta(1-\theta)]^{x_{2}} \cdot(1-\theta)^{2 x_{3}} $$ where \(x_{1}, x_{2}\), and \(x_{3}\) are the number of individuals in the sample who have the AA, Aa, and aa genotypes, respectively. Show that the most powerful test for testing \(H_{0}: \theta=.5\) versus \(H_{\mathrm{a}}: \theta=.8\) rejects the null hypothesis when \(2 x_{1}+x_{2} \geq c\). Is this test UMP for the alternative \(H_{\mathrm{a}}: \theta>.5\) ? Explain. [Note: The fact that the joint distribution of \(X_{1}, X_{2}\), and \(X_{3}\) is multinomial can be used to obtain the value of \(c\) that yields a test with any desired significance level when \(n\) is large.]

A random sample of soil specimens was obtained, and the amount of organic matter \((\%)\) in the soil was determined for each specimen, resulting in the accompanying data (from "Engineering Properties of Soil," Soil Sci., 1998: 93-102). \(\begin{array}{llllllll}1.10 & 5.09 & 0.97 & 1.59 & 4.60 & 0.32 & 0.55 & 1.45 \\\ 0.14 & 4.47 & 1.20 & 3.50 & 5.02 & 4.67 & 5.22 & 2.69 \\ 3.98 & 3.17 & 3.03 & 2.21 & 0.69 & 4.47 & 3.31 & 1.17 \\ 0.76 & 1.17 & 1.57 & 2.62 & 1.66 & 2.05 & & \end{array}\) The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are \(2.481,1.616\), and \(.295\), respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than \(3 \%\) ? Carry out a test of the appropriate hypotheses at significance level 10 by first determining the \(P\)-value. Would your conclusion be different if \(\alpha=.05\) had been used? [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]

Exercise 33 in Chapter 1 gave \(n=26\) observations on escape time (sec) for oil workers in a simulated exercise, from which the sample mean and sample standard deviation are \(370.69\) and \(24.36\), respectively. Suppose the investigators had believed a priori that true average escape time would be at most \(6 \mathrm{~min}\). Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significance level of \(.05\).

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