Answer to Question #291583 in Statistics and Probability for DAM

Question

1)     How do drivers react to sudden large increases in the price of gasoline? To help answer the question, a statistician recorded the speed of cars as they passed a large service station. He recorded the speeds (mph) in the same location after the service station showed that the price of gasoline had risen by 15 cents. Can we conclude that the speeds differ?

Answer 1

Speeds before price increase(1)

$n_1=12\\\bar x_1={\sum x\over n_1}={400\over12}=33.3333$

$s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={13556-13333.3333\over11}=20.2424273$

Speeds after price increase(2)

$n_2=12\\\bar x_2={\sum x\over n_2}={378\over12}=31.5$

$s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={12040-11907\over11}=12.0909091$

Before we go to test the means first we have to test their variability using F-test.

We test,

$H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2$

The test statistic is,

$F_c={s_1^2\over s_2^2}={20.2424273\over12.0909091}=1.67418572$

The table value is,

$F_{\alpha,n_1-1,n_2-1}=F_{0.05,11,11}= 2.81793$ and we reject the null hypothesis if $F_c\gt F_{\alpha,n_1-1,n_2-1}$

Since $F_c=1.67418572\lt F_{0.05,11,11}=2.81793$, we accept the null hypothesis that the population variances for speeds before and after price increase are equal.

Now,

The hypothesis tested are,

$H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\not=\mu_2$ 

The test statistic is,

$t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}$

where $sp^2$ is the pooled sample variance given as,

$sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(11\times20.2424273)+(11\times12.0909091)\over8}={355.6667\over22}=16.1666682$

Therefore,

$t_c={(33.33-31.5)\over \sqrt{16.16667({1\over 12}+{1\over 12})}}={1.8333333\over1.6415}=1.116881$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=12+12-2=22$ degrees of freedom.

The table value is,

$t_{{0.05\over2},22}=t_{0.025,22}=2.074$

The null hypothesis is rejected if $|t_c|\gt t_{0.025,22}.$

Since, $|t_c|=1.116881\lt t_{0.025,22}=2.074,$ we fail to reject the null hypothesis and conclude that there is no sufficient evidence to show that the speeds differ at 5% significance level.