/*! 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 65 A sample of 50 lenses used in ey... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 65

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
Reject the null hypothesis; the true average thickness differs from 3.20 mm.

Step by step solution

01

Define Hypotheses

To determine if the true average thickness of the lenses differs from the desired thickness, we need to set up our hypotheses. The null hypothesis \(H_0\) is that the true average thickness is equal to the desired thickness. The alternative hypothesis \(H_a\) is that the thickness is different from the desired thickness. \(H_0: \mu = 3.20\) and \(H_a: \mu eq 3.20\).
02

Identify Known Values

Identify the values from the problem that will be used in the test. We know the sample mean \(\bar{x} = 3.05\), the sample standard deviation \(s = 0.34\), the sample size \(n = 50\), and the desired level of significance \(\alpha = 0.05\).
03

Calculate Test Statistic

To calculate the test statistic, use the formula for the t-statistic: \(t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\). Plug in the values: \(t = \frac{3.05 - 3.20}{0.34/\sqrt{50}}\).
04

Calculate Test Statistic Result

Compute the test statistic: \(t = \frac{-0.15}{0.0481} \approx -3.12\).
05

Determine Critical Value

Because it's a two-tailed test (since \(H_a\) is \(\mu eq 3.20\)), and \(\alpha = 0.05\), split \(\alpha\) in two to get \(\alpha/2 = 0.025\). Using a t-distribution table or calculator for \(49\) degrees of freedom (\(n-1\)), find the critical t-value \(t_{critical} \approx \pm2.009\).
06

Decision Rule

Compare the calculated t-statistic to the critical t-values. If \(\left| t \right| > \left| t_{critical} \right|\), reject the null hypothesis \(H_0\). Here, \(\left| -3.12 \right| > \left| 2.009 \right|\).
07

Conclusion

Since the calculated t-statistic exceeds the critical values, we reject the null hypothesis \(H_0\). There is sufficient evidence to suggest that the true average thickness of the lenses is different from the desired thickness.

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 91影视!

Key Concepts

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

Understanding the t-statistic
The t-statistic is a fundamental part of hypothesis testing, especially when dealing with small samples or unknown population variances. In this context, it's used to determine how far our sample mean is from the hypothesized population mean, considering the variability in data. The formula for the t-statistic is given by:
\( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \)
where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu_0 \) is the population mean under the null hypothesis,
  • \( s \) is the sample standard deviation,
  • \( n \) is the sample size.
A higher absolute value of the t-statistic indicates that the sample mean is significantly different from the population mean. In this exercise, the t-statistic was calculated to be approximately -3.12, which suggests a substantial deviation.
The Role of the Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a statement that there is no effect or no difference. In hypothesis testing, it serves as a starting point that is tested for validity against the sample data. For this exercise, the null hypothesis is that the true average thickness of lenses is equal to the desired thickness of 3.20 mm, which is expressed as:
\( H_0: \mu = 3.20 \)
The null hypothesis is analyzed against an alternative hypothesis. In this instance, the alternative hypothesis was that the actual average thickness was different from 3.20 mm:
\( H_a: \mu eq 3.20 \)
The primary goal in hypothesis testing is to determine whether there is strong enough evidence from the sample data to reject the null hypothesis.
Understanding the Sample Mean
The sample mean is a measure of central tendency, calculated as the sum of all sample observations divided by the number of observations. It provides a point estimate of the population mean, helping us understand the average outcome in our sample. In the context of the exercise, the sample mean was calculated as 3.05 mm.
This value is crucial in hypothesis testing because it serves as the primary statistic for comparison with the population mean outlined by the null hypothesis. The difference between the sample mean and the hypothesized population mean (3.20 mm) is central to determining if there is a significant discrepancy.
Identifying the Critical Value
The critical value is a threshold used in hypothesis testing to decide whether to accept or reject the null hypothesis. It is determined based on the desired level of significance, \( \alpha \), which reflects the probability of rejecting the null hypothesis when it is true (Type I error). In most statistical tests, \( \alpha \) is set at 0.05.
For this exercise, since it's a two-tailed test, the significance level was split, giving \( \alpha/2 = 0.025 \). Using a t-distribution table, the critical value was found to be approximately \( \pm2.009 \).
The decision to reject or not reject \( H_0 \) is based on comparing the calculated t-statistic with this critical value. Since the absolute t-statistic \( 3.12 \) exceeds \( 2.009 \), the null hypothesis was rejected, indicating a significant difference in thickness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A hot-tub manufacturer advertises that with its heating equipment, a temperature of \(100^{\circ} \mathrm{F}\) can be achieved in at most \(15 \mathrm{~min}\). A random sample of 42 tubs is selected, and the time necessary to achieve a \(100^{\circ} \mathrm{F}\) temperature is determined for each tub. The sample average time and sample standard deviation are \(16.5 \mathrm{~min}\) and \(2.2 \mathrm{~min}\), respectively. Does this data cast doubt on the company's claim? Compute the \(P\)-value and use it to reach a conclusion at level .05.

A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered. a. Does this provide compelling evidence for concluding that more than \(10 \%\) of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of 05 . In reaching your conclusion, what type of error might you have committed? b. If it is really the case that \(15 \%\) of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level .05 test? Answer this question for a sample size of 200 . c. How many plates would have to be tested to have \(\beta(.15)=.10\) for the test of part (a)?

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\bar{x}=52.7\) and a sample standard deviation of \(s=4.8\). The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

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?

A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample \(t\) test. a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied? b. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-2.3\), and \(\alpha=.05 ?\) c. What conclusion is appropriate if the hypotheses of part (a) are tested, \(t=-1.8\), and \(\alpha=.01\) ? d. What should be concluded if the hypotheses of part (a) are tested and \(t=-3.6\) ?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.