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Chapter 9: Problem 3

Let \(\mu_{1}\) denote true average tread life for a premium brand of P205/65R15 radial tire, and let \(\mu_{2}\) denote the true average tread life for an economy brand of the same size. Test \(H_{0}: \mu_{1}-\mu_{2}=5000\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>5000\) at level \(.01\), using the following data: \(m=45, \bar{x}=42,500\), \(s_{1}=2200, n=45, \bar{y}=36,800\), and \(s_{2}=1500\).

Short Answer

Expert verified
Reject the null hypothesis; the premium brand tread life is more than 5000 miles longer than the economy brand.

Step by step solution

01

Identify the Null and Alternative Hypotheses

The null hypothesis is \( H_0: \mu_1 - \mu_2 = 5000 \). The alternative hypothesis is \( H_a: \mu_1 - \mu_2 > 5000 \). This implies we believe the premium brand will have a tread life more than 5000 miles longer than the economy brand.
02

State the Significance Level and Distribution

The significance level \( \alpha \) is 0.01. We will use a t-test for the difference between two means, assuming equal variances, which follows a t-distribution.
03

Calculate the Test Statistic

The formula for the t-test statistic is \[ t = \frac{(\bar{x} - \bar{y}) - D_0}{\sqrt{s_p^2(\frac{1}{m} + \frac{1}{n})}} \]where \(D_0 = 5000\) and \(s_p^2\) is the pooled variance given by \[ s_p^2 = \frac{(m-1)s_1^2 + (n-1)s_2^2}{m+n-2} = \frac{(45-1)\times2200^2 + (45-1)\times1500^2}{45+45-2} \].
04

Compute the Pooled Variance

Calculate \(s_p^2\):\[s_p^2 = \frac{44\times2200^2 + 44\times1500^2}{88} \approx 353611.36\].
05

Calculate the Standard Error of the Difference in Means

Calculate the standard error using \[ \sqrt{s_p^2(\frac{1}{m} + \frac{1}{n})} = \sqrt{353611.36\times(\frac{1}{45} + \frac{1}{45})} \approx 125.131 \].
06

Compute the t-statistic

Substitute the known values to find the t-statistic:\[t = \frac{(42500 - 36800) - 5000}{125.131} = \frac{700}{125.131} \approx 5.60 \].
07

Determine the Critical t-value

For a one-tailed test at \( \alpha = 0.01 \) with \( m+n-2 = 88 \) degrees of freedom, check the t-distribution table to find the critical t-value, which is approximately 2.376.
08

Compare t-statistic with Critical Value and Decide

Since the calculated t-statistic (5.60) is greater than the critical value (2.376), we reject the null hypothesis in favor of the alternative hypothesis.

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Key Concepts

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

t-test
The t-test is a statistical test that helps in comparing the means from two samples. It plays a crucial role in hypothesis testing. This test is particularly useful when we do not know the population standard deviation. In the given exercise, a t-test is used to test if the mean tread life of a premium brand tire exceeds that of an economy brand by a certain amount.
The t-test checks whether the difference between two means is significant or likely due to chance.
  • We assume the difference is equal to a proposed value (5000 in this case).
  • The formula for the t-statistic is \[t = \frac{(\bar{x} - \bar{y}) - D_0}{\sqrt{s_p^2(\frac{1}{m} + \frac{1}{n})}}\]
  • This formula considers the sample means, pooled variance, and the sample sizes.

Once the t-statistic is calculated, we compare it to a critical value from a t-distribution table to make decisions about our hypothesis.
significance level
The significance level, often denoted by \( \alpha \), is a critical probability threshold in hypothesis testing. For our exercise, the significance level is set at 0.01.
This value indicates that there is a 1% chance of rejecting the null hypothesis when it is actually true, also known as a Type I error.
  • A lower significance level, like 0.01, means we require stronger evidence to reject the null hypothesis.
  • When comparing the t-statistic to the critical value, this level helps define the boundaries for decision-making.
  • If our t-statistic exceeds the critical value at the 0.01 level, we can confidently reject the null hypothesis.

    • In simpler terms, the significance level protects against making a false positive decision when testing hypotheses.
pooled variance
Pooled variance is a method used in calculating the variance for two samples assumed to be drawn from populations with equal variances. In the exercise, both tire brands are presumed to have similar variance in tread life, which allows us to pool the variance.
This technique provides a more reliable estimate of variance by taking into account the sample sizes and variances of both groups.
  • The formula for pooled variance is \[ s_p^2 = \frac{(m-1)s_1^2 + (n-1)s_2^2}{m+n-2} \]
  • It helps in accurately calculating the standard error needed for the t-test.
  • Accurate variance calculation ensures that the statistical tests conducted are valid, leading to credible results.

Using pooled variance assumes the variances from two groups being compared are similar, simplifying our calculations.
alternative hypothesis
In hypothesis testing, the alternative hypothesis presents a statement we aim to provide evidence for. In this exercise, the alternative hypothesis \( H_a: \mu_1 - \mu_2 > 5000 \) suggests that the premium brand tires have a significantly longer tread life than the economy brand by more than 5000 miles.
Unlike the null hypothesis stating no effect or a specific condition, the alternative hypothesis is the research hypothesis we wish to verify.
  • It can be one-sided or two-sided (in this case, we use one-sided).
  • We reject the null hypothesis in favor of the alternative when the calculated test statistic exceeds the critical value.
  • Acceptance of the alternative hypothesis implies statistical evidence for a meaningful difference or effect.

    • Understanding the alternative hypothesis drives the direction and purpose of hypothesis testing, focusing our analysis on detecting changes or differences.

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Most popular questions from this chapter

The paper "Quantitative Assessment of Glenohumeral Translation in Baseball Players" (The Amer. J. of Sports Med. 2004: 1711-1715) considered various aspects of shoulder motion for a sample of pitchers and another sample of position players [glenohumeral refers to the articulation between the humerus (ball) and the glenoid (socket)]. The authors kindly supplied the following data on anteroposterior translation (mm), a measure of the extent of anterior and posterior motion, both for the dominant arm and the nondominant arm. \(\begin{array}{lcccc} & \text { Pos Dom Tr Pos ND } & \text { Tr Pit Dom Tr } & \text { Pit ND } & \text { Tr } \\ 1 & 30.31 & 32.54 & 27.63 & 24.33 \\ 2 & 44.86 & 40.95 & 30.57 & 26.36 \\ 3 & 22.09 & 23.48 & 32.62 & 30.62 \\ 4 & 31.26 & 31.11 & 39.79 & 33.74 \\ 5 & 28.07 & 28.75 & 28.50 & 29.84 \\ 6 & 31.93 & 29.32 & 26.70 & 26.71 \\ 7 & 34.68 & 34.79 & 30.34 & 26.45 \\ 8 & 29.10 & 28.87 & 28.69 & 21.49 \\ 9 & 25.51 & 27.59 & 31.19 & 20.82 \\ 10 & 22.49 & 21.01 & 36.00 & 21.75 \\ 11 & 28.74 & 30.31 & 31.58 & 28.32 \\ 12 & 27.89 & 27.92 & 32.55 & 27.22 \\ 13 & 28.48 & 27.85 & 29.56 & 28.86 \\ 14 & 25.60 & 24.95 & 28.64 & 28.58 \\ 15 & 20.21 & 21.59 & 28.58 & 27.15 \\ 16 & 33.77 & 32.48 & 31.99 & 29.46 \\ 17 & 32.59 & 32.48 & 27.16 & 21.26 \\ 18 & 32.60 & 31.61 & & \\ 19 & 29.30 & 27.46 & & \\ \text { mean } & 29.4463 & 29.2137 & 30.7112 & 26.6447 \\ \text { sd } & 5.4655 & 4.7013 & 3.3310 & 3.6679\end{array}\) a. Estimate the true average difference in translation between dominant and nondominant arms for pitchers in a way that conveys information about reliability and precision, and interpret the resulting estimate. b. Repeat (a) for position players. c. The authors asserted that "pitchers have greater difference in side-to-side anteroposterior translation of their shoulders compared with position players." Do you agree? Explain.

The article "Pine Needles as Sensors of Atmospheric Pollution" (Environ. Monitoring, 1982: 273-286) reported on the use of neutron-activity analysis to determine pollutant concentration in pine needles. According to the article's authors, "These observations strongly indicated that for those elements which are determined well by the analytical procedures, the distribution of concentration is lognormal. Accordingly, in tests of significance the logarithms of concentrations will be used." The given data refers to bromine concentration in needles taken from a site near an oil-fired steam plant and from a relatively clean site. The summary values are means and standard deviations of the log-transformed observations. \begin{tabular}{lccc} Site & Sample Size & Mean Log Concentration & SD of Log Concentration \\ \hline Steam plant & 8 & \(18.0\) & \(4.9\) \\ Clean & 9 & \(11.0\) & \(4.6\) \\ \hline \end{tabular} Let \(\mu_{1}^{*}\) be the true average log concentration at the first site, and define \(\mu_{2}^{*}\) analogously for the second site. a. Use the pooled \(t\) test (based on assuming normality and equal standard deviations) to decide at significance level .05 whether the two concentration distribution means are equal. b. If \(\sigma_{1}^{*}\) and \(\sigma_{2}^{*}\) (the standard deviations of the two log concentration distributions) are not equal, would \(\mu_{1}\) and \(\mu_{2}\) (the means of the concentration distributions) be the same if \(\mu_{1}^{*}=\mu_{2}^{*}\) ? Explain your reasoning.

Wait staff at restaurants have employed various strategies to increase tips. An article in the Sept. 5, 2005, New Yorker reported that "In one study a waitress received \(50 \%\) more in tips when she introduced herself by name than when she didn't." Consider the following (fictitious) data on tip amount as a percentage of the bill: \(\begin{array}{lrll}\text { Introduction: } & m=50 & \bar{x}=22.63 & s_{1}=7.82 \\ \text { No introduction: } & n=50 & \bar{y}=14.15 & s_{2}=6.10\end{array}\) Does this data suggest that an introduction increases tips on average by more than \(50 \%\) ? State and test the relevant hypotheses. [Hint: Consider the parameter \(\theta=\mu_{1}-1.5 \mu_{2}\).]

An experiment to determine the effects of temperature on the survival of insect eggs was described in the article "Development Rates and a Temperature- Dependent Model of Pales Weevil" (Environ. Entomology, 1987:956-962). At \(11^{\circ} \mathrm{C}, 73\) of 91 eggs survived to the next stage of development. At \(30^{\circ} \mathrm{C}, 102\) of 110 eggs survived. Do the results of this experiment suggest that the survival rate (proportion surviving) differs for the two temperatures? Calculate the \(P\)-value and use it to test the appropriate hypotheses.

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. \begin{tabular}{lrr} & \(\mathbf{A}\) & \(\mathbf{B}\) \\ \hline \(26-<30\) & 6 & 4 \\ \(30-<34\) & 12 & 9 \\ \(34-<38\) & 15 & 19 \\ \(38-<42\) & 7 & 10 \\ & \(m=40\) & \(m=42\) \\ \hline \end{tabular} Compute a \(95 \%\) CI for the difference between the true proportions of all specimens of alloys \(\mathrm{A}\) and \(\mathrm{B}\) that have an ultimate strength of at least \(34 \mathrm{ksi}\).
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