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10.36 \text { A large automobile manufacturing company is } trying to decide whether to purchase brand \(A\) or brand \(B\) tires for its new models. To help arrive at a decision, an experiment is conducted using 12 of each brand. The tires are run until they wear out. The results are $$\begin{aligned}\text { Brand } A: & x_{1}=37,900 \text { kilometers, } \\ & s_{1}=5,100 \text { kilometers. } \\\\\text { Brand B: } \quad x_{1} &=39,800 \text { kilometers, } \\\& s_{2}=5,900 \text { kilometers. }\end{aligned}$$ Test the hypothesis that there is no difference in the average wear of 2 brands of tires. Assume the populations to be approximately normally distributed with equal variances. Use a P-value.

Short Answer

Expert verified
Result of test will be based on P-value. If P-value is less than or equal to 0.05, then we reject the null hypothesis and conclude there's existing difference in the average wear of two brands of tires. If P-value is greater than 0.05, we fail to reject the null hypothesis and conclude that there is not enough evidence to support a difference in the average wear of the two brands of tires.

Step by step solution

01

State the Hypotheses

The null hypothesis \(H_0\) states that there is no difference in the average wear of the two brands of tires. The alternative hypothesis \(H_1\) states that there is a difference in the average wear. In mathematical terms, these hypotheses are represented as \(H_0: \mu_A = \mu_B\) and \(H_1: \mu_A \neq \mu_B\).
02

Calculate the Test Statistic

The formula for the test statistic in a two independent samples t-test is: \(t = \frac{(x_1 - x_2) - (µ_x1 - µ_x2)}{\sqrt{(s^2/n1) + (s^2/n2)}}\) where \(x_1\) and \(x_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, and \(n_1\) and \(n_2\) are the number of tires tested of each brand. Substituting the given values, calculate the t value.
03

Determine the Degrees of Freedom

The degrees of freedom for a two independent samples t-test equals to the sum of the two sample sizes minus 2. The formula here is \(df = n_1 + n_2 - 2\). Substituting the values \(n_1 = 12\) and \(n_2 = 12\), calculate the degrees of freedom.
04

Calculate the P-value

The P-value is the probability that you have falsely rejected the null hypothesis. Look up the t-value got in Step 2 in a t-distribution table using the degrees of freedom determined in Step 3 to find the P-value. Since we are looking at a two-tailed test (because the alternative hypothesis states \(µ_A \neq µ_B\)), make sure to double the P-value obtained from the t-table.
05

Make A Decision

Compare the p-value to the significance level. If the P-value is less than or equal to the significance level (usually 0.05), we reject the null hypothesis. If the P-value is greater than the significance level, we fail to reject the null hypothesis.

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Key Concepts

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

T-Test
A t-test is a statistical tool used to determine if there is a significant difference between the means of two groups, which may be related in certain features. It's particularly useful when dealing with small sample sizes (<30) where the central limit theorem doesn't fully apply, hence the normal distribution isn't a great fit. In our tire example, a t-test helps us decide if the wear difference between Brand A and Brand B is statistically meaningful, or if it occurred by random chance. The calculation revolves around the sample means, standard deviations, and sizes, bringing us to the question: are the observed average kilometers until wear out for each tire brand systematically different or not.

Understanding this concept is crucial when analyzing experiments where comparing two means is needed under the uncertainty of small samples. In business decisions like choosing a tire supplier, the correct application of the t-test can lead to better choices backed by statistical evidence.
Statistical Significance
The notion of statistical significance plays a pivotal role in hypothesis testing. It provides a criterion for deciding whether the observed effect—such as the difference between the average tire wear—could be due to chance. If a result is statistically significant, it means that the likelihood of the event occurring by chance is low enough that we can confidently attribute it to the factor being tested—in our case, the tire brand.

The conventional threshold for declaring statistical significance is a p-value of 0.05 or less. This means there is less than a 5% chance the observed difference in tire wear occurred in a world where, in reality, the two brands are equivalent. Statistically significant results prompt us to reject the null hypothesis, influencing decisions such as selecting a tire brand for an entire fleet of vehicles.
Sample Mean and Standard Deviation
In any statistical analysis, the sample mean represents the average outcome, whereas the standard deviation reflects how much the individual data points vary from the mean. For Brand A and B tires, the means are the average kilometers until wear—37,900 km for Brand A and 39,800 km for Brand B. The standard deviations—5,100 km for Brand A and 5,900 km for Brand B—show the variability within each brand's tire life.

Both these statistics are used to compute the t-test and craft a conclusion about the tire longevity. They give us a sense of what to expect on average (mean) and how consistent that expectation is (standard deviation). A small standard deviation relative to the mean suggests that we can predict tire wear more precisely, which is a valuable consideration for our vehicle manufacturer.
P-Value
The p-value, a term that sometimes mystifies, is actually a key concept in hypothesis testing. It is the probability of obtaining test results at least as extreme as the ones observed, assuming the null hypothesis is true. In simple terms, it tells us how surprised we should be by the data if there was truly no difference between the tire brands.

In our scenario, a low p-value would suggest that the observed difference in kilometers of Brand A and Brand B tires is not just due to random variation—it's statistically significant. Conversely, a high p-value indicates that any difference could easily have arisen by chance. Therefore, when the p-value falls below our significance level (commonly 0.05), we have grounds to question the null hypothesis, giving us the statistical confidence needed to make well-informed decisions.

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Most popular questions from this chapter

Suppose that all allergist wishes to test the hy pothesis that at least \(30 \%\) of the public is allergic to some cheese products. Explain how the allergist could commit (a) a type I error (b) a type II error.

The average height of females in the freshman class of a certain college has been 162.5 centimeters with a standard deviation of 6.9 centimeters. Is there reason to believe that there has been a change in the average height if a random sample of 50 females in the present freshman class has an average height of 165.2 centimeters? Use a P-value in your conclusion. Assume the standard deviation remains the same

A large manufacturing hirm is being charged with discrimination in its hiring practices. (a) What hypothesis is being tested if a jury commits a type I error by dueling the firm guilty? (b) What hypothesis is being tested if a jury commits a type Il error by finding the firm guilty?

It is claimed that an automobile is driven on the average more than 20,000 kilometers per year. To test this claim, a random sample of 100 automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 23,500 kilometers and a standard deyiation of 3900 kilometers? Use a P-value in your conclusion.

An urban community would like to show that the incidence of breast cancer is higher than in a nearbyrural area. (PCB levels were found to be higher in the soil of the urban community.) If it is found that 20 of 200 adult women in the urban community have breast cancer and 10 of 150 adult women in the rural commus nity have breast cancer, can we conclude at the 0.05 level of significance that breast cancer is more prevalent in the urban community?

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