Question #291583

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?


1
Expert's answer
2022-01-31T15:00:22-0500

Speeds before price increase(1)

n1=12xˉ1=xn1=40012=33.3333n_1=12\\\bar x_1={\sum x\over n_1}={400\over12}=33.3333

s12=x2(x)2n1n11=1355613333.333311=20.2424273s_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)

n2=12xˉ2=xn2=37812=31.5n_2=12\\\bar x_2={\sum x\over n_2}={378\over12}=31.5

s22=x2(x)2n2n21=120401190711=12.0909091s_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,

H0:σ12=σ22vsH1:σ12σ22H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2

The test statistic is,

Fc=s12s22=20.242427312.0909091=1.67418572F_c={s_1^2\over s_2^2}={20.2424273\over12.0909091}=1.67418572

The table value is,

Fα,n11,n21=F0.05,11,11=2.81793F_{\alpha,n_1-1,n_2-1}=F_{0.05,11,11}= 2.81793 and we reject the null hypothesis if Fc>Fα,n11,n21F_c\gt F_{\alpha,n_1-1,n_2-1}

Since Fc=1.67418572<F0.05,11,11=2.81793F_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,

H0:μ1=μ2vsH1:μ1μ2H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\not=\mu_2 

The test statistic is,

tc=(xˉ1xˉ2)sp2(1n1+1n2)t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}

where sp2sp^2 is the pooled sample variance given as,

sp2=(n11)s12+(n21)s22n1+n22=(11×20.2424273)+(11×12.0909091)8=355.666722=16.1666682sp^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,

tc=(33.3331.5)16.16667(112+112)=1.83333331.6415=1.116881t_c={(33.33-31.5)\over \sqrt{16.16667({1\over 12}+{1\over 12})}}={1.8333333\over1.6415}=1.116881

tct_c is compared with the table value at α=0.05\alpha=0.05 with n1+n22=12+122=22n_1+n_2-2=12+12-2=22 degrees of freedom.

The table value is,

t0.052,22=t0.025,22=2.074t_{{0.05\over2},22}=t_{0.025,22}=2.074

The null hypothesis is rejected if tc>t0.025,22.|t_c|\gt t_{0.025,22}.

Since, tc=1.116881<t0.025,22=2.074,|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.


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