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Chapter 9: Problem 19

Gasoline Prices A random sample of monthly gasoline prices was taken from 2011 and from 2015. The samples are shown. Using \(\alpha=0.01,\) can it be concluded that gasoline cost more in 2015? Use the \(P\) -value method. $$ \frac{2011}{2015} | \begin{array}{cccccc}{2011} & {2.02} & {2.47} & {2.50} & {2.70} & {3.13} & {2.56} \\ \hline 2015 & {2.36} & {2.46} & {2.63} & {2.76} & {3.00} & {2.85} & {2.77}\end{array} $$

Short Answer

Expert verified
Gasoline prices in 2015 were not significantly higher than in 2011 at \(\alpha = 0.01\).

Step by step solution

01

State the Hypotheses

First, define the null and alternative hypotheses. The null hypothesis \(H_0\) states that there is no difference in gasoline prices between 2011 and 2015. The alternative hypothesis \(H_1\) states that gasoline prices in 2015 are greater than those in 2011. Formally, \(H_0: \mu_1 = \mu_2\) (mean price in 2011 is equal to mean price in 2015) and \(H_1: \mu_1 < \mu_2\) (mean price in 2015 is greater than mean price in 2011).
02

Calculate the Sample Means and Standard Deviations

Compute the sample means \(\bar{x}_1\) and \(\bar{x}_2\) for 2011 and 2015 respectively. Calculate the sample standard deviations \(s_1\) and \(s_2\). For 2011: Mean \(\bar{x}_1 = \frac{2.02 + 2.47 + 2.50 + 2.70 + 3.13 + 2.56}{6} = 2.5633\) and standard deviation \(s_1 = 0.3947\).For 2015: Mean \(\bar{x}_2 = \frac{2.36 + 2.46 + 2.63 + 2.76 + 3.00 + 2.85 + 2.77}{7} = 2.6914\) and standard deviation \(s_2 = 0.2157\).
03

Determine the Test Statistic

Since variances are unequal and sample sizes are small, use the t-test for unequal variances. The test statistic \(t\) is calculated as:\[ t = \frac{(\bar{x}_2 - \bar{x}_1)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Substitute values: \[ t = \frac{2.6914 - 2.5633}{\sqrt{\frac{0.3947^2}{6} + \frac{0.2157^2}{7}}} = 0.8278 \]
04

Find the Degrees of Freedom

Use the formula for degrees of freedom for two samples:\[ df \approx \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \]Plug in known values to find approximately \(df = 10.89\). For practicality, use \(df = 10\).
05

Determine the P-value

Using a t-distribution table or calculator, calculate the p-value corresponding to the test statistic \(t = 0.8278\) with \(10\) degrees of freedom. The one-tailed p-value is approximately \(p = 0.213\).
06

Make a Decision

Compare the p-value to the significance level \(\alpha = 0.01\). Since \(p > \alpha\), do not reject the null hypothesis. There is insufficient evidence to conclude that the gasoline prices in 2015 are statistically significantly higher than in 2011.

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Key Concepts

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

t-test for unequal variances
In statistics, when comparing the means of two independent samples, it's common to use a t-test. But when sample variances differ significantly and sample sizes are small, the typical t-test assumptions may not hold. Here, a 't-test for unequal variances' or Welch's t-test comes in handy. This test is an adaptation that doesn't assume equal variances or equal sample sizes.

For this test, you'll calculate a test statistic, which helps determine if observed differences between sample means are due to chance. Specifically, the formula for the t-statistic when variances are unequal is:
  • Calculate the difference between the two sample means.
  • Divide by the square root of the sum of the squared sample deviations divided by their respective sizes.
By addressing unequal variances, Welch's t-test provides a more reliable measure for certain data sets, helping make sound statistical conclusions.
p-value method
In hypothesis testing, the p-value method allows for determining the significance of results from your data. A p-value helps us weigh evidence against a null hypothesis (that there is no effect or difference).

You start by calculating the p-value associated with your test statistic, which reflects the probability of observing effects as extreme as those in your data, assuming the null hypothesis is true.
  • If the p-value is less than the chosen significance level (commonly 0.05 or 0.01), there is strong evidence against the null, prompting its rejection.
  • If it's higher, the evidence is not sufficient to reject the null, hence, it is retained.
P-values serve as a practical tool, aiding researchers to objectively interpret the strength of their results without subjective bias.
degrees of freedom
Degrees of freedom in the context of a t-test refer to the number of values in a calculation that are free to vary. This concept influences the shape of the sampling distribution, hence affecting the probability calculations in hypothesis testing.

In a t-test for unequal variances, degrees of freedom are calculated using a more complex formula:
  • This formula takes into account the variance and sample size of each sample.
  • It approximates the number of independent values that affect the statistical computation.
For practical purposes, degrees of freedom in such cases might not be a whole number. Round it to the nearest integer as a pragmatic step for referencing tables or software that may require this input for further calculations.
significance level
A significance level, commonly denoted by the Greek letter alpha (\(\alpha\)), is a threshold for deciding statistical significance. It is a predetermined level at which you decide how much risk you're willing to take in mistakenly rejecting a true null hypothesis.

In many studies, \(\alpha\) is set at 0.05 or 0.01. This setting influences the error rate:
  • An alpha of 0.05 means you accept a 5% chance of concluding an effect exists, when in fact it doesn't.
  • Choosing a more stringent alpha, like 0.01, reduces the risk of this error, thus demanding stronger evidence to reject the null hypothesis.
The significance level is integral, guiding the threshold for p-value comparison and ultimately steering the conclusions of hypothesis tests.

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Most popular questions from this chapter

Instead of finding the mean of the differences between \(X_{1}\) and \(X_{2}\) by subtracting \(X_{1}-X_{2},\) you can find it by finding the means of \(X_{1}\) and \(X_{2}\) and then subtracting the means. Show that these two procedures will yield the same results.

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. High School Graduation Rates The overall U.S. public high school graduation rate is \(73.4 \% .\) For Pennsylvania it is \(83.5 \%\) and for Idaho \(80.5 \%-\mathrm{a}\) difference of \(3 \%\). Random samples of 1200 students from each state indicated that 980 graduated in Pennsylvania and 940 graduated in Idaho. At the 0.05 level of significance, can it be concluded that there is a difference in the proportions of graduating students between the states?

Explain the difference between testing a single mean and testing the difference between two means.

What three assumptions must be met when you are using the \(z\) test to test differences between two means when \(\sigma_{1}\) and \(\sigma_{2}\) are known?

For Exercises 9 through \(24,\) perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Noise Levels in Hospitals In a hospital study, it was found that the standard deviation of the sound levels from 20 randomly selected areas designated as "casualty doors" "was 4.1 dB A and the standard deviation of 24 randomly selected areas designated as operating theaters was 7.5 dB A. At \(\alpha=0.05,\) can you substantiate the claim that there is a difference in the standard deviations?
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