/*! 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 8 Let \(x\) be a random variable t... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x\) be a random variable that represents assembly times for the Ford Taurus. The Wall Street Journal reported that the average assembly time is \(\mu=38\) hours. A modification to the assembly procedure has been made. Experience with this new method indicates that \(\sigma=1.2\) hours. It is thought that the average assembly time may be reduced by this modification. A random sample of 47 new Ford Taurus automobiles coming off the assembly line showed the average assembly time of the new method to be \(\bar{x}=37.5\) hours. Does this indicate that the average assembly time has been reduced? Use \(\alpha=0.01\).

Short Answer

Expert verified
Yes, the average assembly time has been statistically reduced.

Step by step solution

01

State the Hypotheses

To determine if the average assembly time has been reduced, we set up our null and alternative hypotheses. The null hypothesis is that the new average assembly time is equal to the old average time, \( \mu = 38 \) hours. The alternative hypothesis is that the new average assembly time is less than the old average time, \( \mu < 38 \) hours.
02

Establish the Significance Level

The significance level \( \alpha \) is given as 0.01. This means we are looking for evidence strong enough to suggest a reduction in assembly time, and we will reject the null hypothesis if the p-value is less than 0.01.
03

Find the Standard Error

Calculate the standard error (SE) using the formula \( \text{SE} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 1.2 \) and \( n = 47 \). This gives \( \text{SE} = \frac{1.2}{\sqrt{47}} \approx 0.175 \).
04

Calculate the Test Statistic

We calculate the test statistic using the formula \( \text{z} = \frac{\bar{x} - \mu}{\text{SE}} \), where \( \bar{x} = 37.5 \) and \( \mu = 38 \). So, \( z = \frac{37.5 - 38}{0.175} \approx -2.857 \).
05

Determine the P-value

Using the standard normal distribution table, find the p-value corresponding to \( z = -2.857 \). The p-value for \( z = -2.857 \) is approximately 0.0021.
06

Compare the P-value to the Significance Level

Since the p-value (0.0021) is less than the significance level (0.01), we reject the null hypothesis. This suggests that the modification has statistically reduced the average assembly time.

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Key Concepts

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

Null Hypothesis
The first step in hypothesis testing involves setting up a null hypothesis. The null hypothesis is essentially a statement reflecting no change or status quo. In the context of assembly times for the Ford Taurus, the null hypothesis depicts that the new average assembly time remains the same as the old average time. This is represented as:
  • \( H_0: \mu = 38 \) hours
By stating that there is no effect or no difference, the null hypothesis serves as a benchmark against which we compare our observed data. It's what we assume to be true until we have enough evidence to suggest otherwise.
Rejecting a null hypothesis in hypothesis testing usually means that the hypothesis does not fit the observed data and an alternative explanation may be valid. It's important to handle this concept correctly since it forms the foundation of statistical decision-making.
Alternative Hypothesis
What if the null hypothesis isn't representative of reality? That's where the alternative hypothesis comes in. In our situation, the alternative hypothesis suggests that the new procedure has indeed reduced the average assembly time for the Ford Taurus.
  • \( H_a: \mu < 38 \) hours
The alternative hypothesis is what you aim to prove in a test. It stands opposite to the null hypothesis. When testing this hypothesis, we are looking for statistical evidence to support the claim that the average time is less than 38 hours.
It's crucial because it provides the direction for your hypothesis test, indicating what you suspect might be true based on observations or theories. In this particular case, you're aiming to demonstrate that the modification in the assembly process is effective in reducing time.
Significance Level
The significance level, often denoted as \( \alpha \), plays a pivotal role in hypothesis testing. It specifies the threshold for rejecting the null hypothesis. A lower \( \alpha \) value means stricter evidence is required to reject the null hypothesis. For this exercise, the significance level is set at 0.01.
  • \( \alpha = 0.01 \)
This means that there's only a 1% risk of concluding that the average assembly time has reduced due to the modification when in fact it has not. A common analogy is considering it a risk of making a 'false positive' decision in testing.
The chosen significance level must balance between being too strict, possibly missing out on important findings, and being too lenient, potentially finding effects that do not actually exist. It reflects how confident you need to be about your results being statistically significant.
Standard Error
Standard error (SE) is a critical concept that serves as a measure of the statistical accuracy of an estimate. It effectively tells us about the variability of the sample mean in relation to the true population mean. In our assembly time scenario, the SE is calculated using the formula:
  • \( \text{SE} = \frac{\sigma}{\sqrt{n}} \)
Where \(\sigma = 1.2\) hours and \(n = 47\). This calculation results in a standard error of approximately 0.175.
Understanding the standard error helps in assessing how much the sample mean (\(\bar{x}\) ) deviates from the true population mean (\(\mu\) ). A smaller SE suggests that your sample mean is a precise estimator of the population mean. It also feeds into calculating the z-value, which is integral for determining the p-value during hypothesis testing.

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Most popular questions from this chapter

Based on information from the Rocky Mountain News, a random sample of \(n_{1}=12\) winter days in Denver gave a sample mean pollution index of \(\bar{x}_{1}=43\). Previous studies show that \(\sigma_{1}=21\). For Englewood (a suburb of Denver), a random sample of \(n_{2}=14\) winter days gave a sample mean pollution index of \(\bar{x}_{2}=36\). Previous studies show that \(\sigma_{2}=15\). Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a \(1 \%\) level of significance.

In environmental studies, sex ratios are of great importance. Wolf society, packs, and ecology have been studied extensively at different locations in the U.S. and foreign countries. Sex ratios for eight study sites in northern Europe are shown below (based on The Wolf by L. D. Mech, University of Minnesota Press). \(\begin{array}{lcc} \hline \text { Location of Wolf Pack } & \text { \% Males (Winter) } & \text { \% Males (Summer) } \\ \hline \text { Finland } & 72 & 53 \\ \text { Finland } & 47 & 51 \\ \text { Finland } & 89 & 72 \\ \text { Lapland } & 55 & 48 \\ \text { Lapland } & 64 & 55 \\ \text { Russia } & 50 & 50 \\ \text { Russia } & 41 & 50 \\ \text { Russia } & 55 & 45 \\ \hline \end{array}\) It is hypothesized that in winter, "loner" males (not present in summer packs) join the pack to increase survival rate. Use a \(5 \%\) level of significance to test the claim that the average percentage of males in a wolf pack is higher in winter.

A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated "not important" and 6 designated "extremely important." \(\mathrm{A}\) random sample of \(n_{1}=46\) adults were asked about fishing as an outdoor activity. The mean response was \(\bar{x}_{1}=4.9 .\) Another random sample of \(n_{2}=51\) adults were asked about camping as an outdoor activity. For this group, the mean response was \(\bar{x}_{2}=4.3 .\) From previous studies, it is known that \(\sigma_{1}=1.5\) and \(\sigma_{2}=1.2 .\) Does this indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a \(5 \%\) level of significance. Note: A Likert scale usually has to do with approval of or agreement with a statement in a questionnaire. For example, respondents are asked to indicate whether they "strongly agree," "agree," "disagree," or "strongly disagree" with the statement.

In the following data pairs, A represents the cost of living index for utilities and \(B\) represents the cost of living index for transportation. The data are paired by metropolitan areas in the United States. A random sample of 46 metropolitan areas gave the following information. (Reference: Statistical Abstract of the United States, 121 st edition.) \(\begin{array}{c|ccccccccc} \hline A: & 90 & 84 & 85 & 106 & 83 & 101 & 89 & 125 & 105 \\ \hline B: & 100 & 91 & 103 & 103 & 109 & 109 & 94 & 114 & 113 \\ \hline A: & 118 & 133 & 104 & 84 & 80 & 77 & 90 & 92 & 90 \\ \hline B: & 120 & 130 & 117 & 109 & 107 & 104 & 104 & 113 & 101 \\ \hline \hline A: & 106 & 95 & 110 & 112 & 105 & 93 & 119 & 99 & 109 \\ \hline B: & 96 & 109 & 103 & 107 & 103 & 102 & 101 & 86 & 94 \\ \hline A: & 109 & 113 & 90 & 121 & 120 & 85 & 91 & 91 & 97 \\ \hline B: & 88 & 100 & 104 & 119 & 116 & 104 & 121 & 108 & 86 \\ \hline A: & 95 & 115 & 99 & 86 & 88 & 106 & 80 & 108 & 90 & 87 \\ \hline B: & 100 & 83 & 88 & 103 & 94 & 125 & 115 & 100 & 96 & 127 \\ \hline \end{array}\) i. Let \(d\) be the random variable \(d=A-B\). Use a calculator to verify that \(\bar{d} \approx-5.739\) and \(s_{d} \approx 15.910 .\) ii. Do the data indicate that the U.S. population mean cost of living index for utilities is less than that for transportation in these areas? Use \(\alpha=0.05\).

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