/*! 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 21 A manufacturing company receives... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A manufacturing company receives a shipment of 1,000 bolts of nominal shear strength \(4,350 \mathrm{lb}\). A quality control inspector selects five bolts at random and measures the shear strength of each. The data are: $$ 4,3204,2904,3604,3504,320 $$ a. Assuming a normal distribution of shear strengths, test the null hypothesis that the mean shear strength of all bolts in the shipment is \(4,350 \mathrm{lb}\) versus the alternative that it is less than \(4,350 \mathrm{lb}\), at the \(10 \%\) level of significance. b. Estimate the \(p\) -value (observed significance) of the test of part (a). c. Compare the \(p\) -value found in part (b) to \(\alpha=0.10\) and make a decision based on the \(p\) -value approach. Explain fully.

Short Answer

Expert verified
The mean shear strength is less than 4,350 lb at the 10% level.

Step by step solution

01

Calculate sample mean and standard deviation

First, we need to compute the sample mean \( \bar{x} \) and standard deviation \( s \) of the measured shear strengths. The sample data are \( 4320, 4290, 4360, 4350, 4320 \). To find the mean: \[ \bar{x} = \frac{4320 + 4290 + 4360 + 4350 + 4320}{5} = \frac{21640}{5} = 4328 \mathrm{lb} \]To calculate the standard deviation, first compute the deviations from the mean, square them, sum them, then divide by the number of observations minus 1, and finally, take the square root:\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{(4320-4328)^2 + (4290-4328)^2 + (4360-4328)^2 + (4350-4328)^2 + (4320-4328)^2}{5-1}} \]\[ s = \sqrt{\frac{64 + 1444 + 1024 + 484 + 64}{4}} = \sqrt{770} \approx 27.75 \mathrm{lb} \]
02

Set up the hypothesis and calculate t-statistic

We set the null hypothesis \( H_0: \mu = 4350 \) and the alternative hypothesis \( H_1: \mu < 4350 \). The t-statistic is calculated as:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{4328 - 4350}{27.75/\sqrt{5}} \approx \frac{-22}{12.41} \approx -1.77 \]
03

Determine critical t-value and compare

Using a t-table or calculator, we find the critical t-value for \( \alpha = 0.10 \) and \( df = n - 1 = 4 \). The critical value is approximately \( t_{crit} = -1.533 \).Since the calculated t-statistic \( t = -1.77 \) is less than \( t_{crit} \), we reject the null hypothesis.
04

Estimate the p-value

To estimate the p-value for the t-statistic \( t = -1.77 \) with \( 4 \) degrees of freedom from a t-distribution table or software, the p-value is calculated to be approximately \( p = 0.0822 \).
05

Make a decision based on p-value

Compare the p-value (0.0822) to the significance level \( \alpha = 0.10 \). Since \( p < \alpha \), we reject the null hypothesis. This indicates there is enough evidence to conclude that the mean shear strength is less than \( 4350 \mathrm{lb} \) at the 10% significance level.

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 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution forms the backbone of statistics and data analysis. It is a continuous probability distribution characterized by its bell-shaped curve. In a normal distribution, most of the data points cluster around the mean, and as you move away from the mean, the occurrence of data points reduces symmetrically on both sides.
One key feature of the normal distribution is its symmetry about the mean. This means that half of the data lies below the mean, and half lies above. The standard deviation in a normal distribution measures how spread out the numbers are. A smaller standard deviation means the data points are close to the mean, while a larger standard deviation indicates a wider spread.
When performing hypothesis tests, like in the given problem where we test bolt shear strength, assuming a normal distribution simplifies the calculations. It allows for the use of statistical tests, such as the t-test, to make inferences about the population mean based on sample data.
t-statistic
The t-statistic plays a crucial role in hypothesis testing when dealing with sample data. It helps compare the sample mean to a hypothesized population mean. The formula for the t-statistic is:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Let's break this down:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the hypothesized population mean under the null hypothesis.
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
The numerator \(\bar{x} - \mu\) represents the difference between the sample mean and the hypothesized mean.
The denominator \(s/\sqrt{n}\) is the standard error, which adjusts the standard deviation based on the sample size.
In the exercise, we calculated the t-statistic to determine whether the mean shear strength of bolts is significantly less than the specified strength.
p-value
The p-value, or probability value, is a measure that helps us decide whether to reject the null hypothesis. It's an essential part of hypothesis testing, indicating the probability of obtaining test results at least as extreme as the observed results, assuming the null hypothesis is true.
In practical terms, a smaller p-value suggests stronger evidence against the null hypothesis. If the p-value is less than the significance level (commonly set at 0.05 or 0.10), we reject the null hypothesis.
In our bolt testing exercise, a p-value of approximately 0.0822 was found. Since this p-value is less than the significance level of 0.10, it suggests that the test results provide sufficient evidence to reject the null hypothesis, indicating that the mean shear strength might indeed be less than 4,350 lb.
Significance Level
The significance level, often denoted by \(\alpha\), is a threshold we set to decide when to reject the null hypothesis. It's a measure of the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.Common significance levels are 0.05, 0.01, and 0.10. An \(\alpha\) level of 0.05 means we are willing to accept a 5% chance of incorrectly rejecting the true null hypothesis.
Setting a significance level is subjective and depends on the context of the test. In this exercise, a 10% significance level is used, meaning the inspector is comfortable with a 10% risk of concluding that the mean shear strength is less than 4,350 lb when, in fact, it might not be.
Understanding the significance level helps interpret the p-value correctly. It serves as a benchmark to help make informed decisions in statistical hypothesis testing.

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

Perform the indicated test of hypotheses, based on the information given. a. Test \(H_{0: \mu}=105\) vs. Ha: \(\mu>105 @ \alpha=0.05, \sigma\) unknown, \(n=30, x-=108, s=7.2\) b. Test \(H 0: \mu=21.6\) vs. Ha: \(\mu<21.6 @ \alpha=0.01, \sigma\) unknown, \(n=78, x-=20.5, s=3.9\) c. Test \(H 0: \mu=-375\) VS. Ha: \(\mu \neq-375 @ \alpha=0.01, \sigma=18.5, n=31, x-=-388, s=18.0\)

Ice cream is legally required to contain at least \(10 \%\) milk fat by weight. The manufacturer of an economy ice cream wishes to be close to the legal limit, hence produces its ice cream with a target proportion of 0.106 milk fat. A sample of five containers yielded a mean proportion of 0.094 milk fat with standard deviation 0.002. Test the null hypothesis that the mean proportion of milk fat in all containers is 0.106 against the alternative that it is less than \(0.106,\) at the \(10 \%\) level of significance. Assume that the proportion of milk fat in containers is normally distributed. Hint: This exercise could have been presented in an earlier section.

The average household size in a certain region several years ago was 3.14 persons. A sociologist wishes to test, at the \(5 \%\) level of significance, whether it is different now. Perform the test using the information collected by the sociologist: in a random sample of 75 households, the average size was 2.98 persons, with sample standard deviation 0.82 person.

In the past, \(68 \%\) of a garage's business was with former patrons. The owner of the garage samples 200 repair invoices and finds that for only 114 of them the patron was a repeat customer. a. Test whether the true proportion of all current business that is with repeat customers is less than \(68 \%,\) at the \(1 \%\) level of significance. b. Compute the observed significance of the test.

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \cdot \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average July temperature in a region historically has been \(74.5^{\circ} \mathrm{F}\). Perhaps it is higher now. b. The average weight of a female airline passenger with luggage was 145 pounds ten years ago. The FAA believes it to be higher now. c. The average stipend for doctoral students in a particular discipline at a state university is \(\$ 14,756\). The department chairman believes that the national average is higher. d. The average room rate in hotels in a certain region is \(\$ 82.53 .\) A travel agent believes that the average in a particular resort area is different. e. The average farm size in a predominately rural state was 69.4 acres. The secretary of agriculture of that state asserts that it is less today.

See all solutions

Recommended explanations on Math Textbooks

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.