Question

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

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

Step-by-step solution

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).

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 \(s_2 = 0.2157\).

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: \[ t = \frac{2.6914 - 2.5633}{\sqrt{\frac{0.3947^2}{6} + \frac{0.2157^2}{7}}} = 0.8278 \]

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 \(df = 10\).

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 \(p = 0.213\).

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.

Key concepts

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.