Question

an industrial plant wants to determine which of two types of fuel, electric or gas, is more cost - efficient (measured in cost per unit of energy). independent random samples were taken of plants using electricity and plants using gas. these samples consisted of 7 plants using electricity, which had a mean cost per unit of $44.58 and standard deviation of $7.92, and 9 plants using gas, which had a mean of $54.40 and standard deviation of $7.87. assume that the populations of costs per unit are normally distributed for each type of fuel, and assume that the variances of these populations are equal. can we conclude, at the 0.10 level of significance, that $mu_1$, the mean cost per unit for plants using electricity, differs from $mu_2$, the mean cost per unit for plants using gas? perform a two - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places and round your answers as specified in the table. (if necessary, consult a list of formulas.) (a) state the null hypothesis $h_0$ and the alternative hypothesis $h_1$. $h_0$: $h_1$: (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) (d) find the two critical values at the 0.10 level of significance. (round to three or more decimal places.) and (e) can we conclude that the mean cost per unit for plants using electricity differs from the mean cost per unit for plants using gas? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the means are equal, i.e., $H_0:\mu_1-\mu_2 = 0$. The alternative hypothesis $H_1$ for a two - tailed test is that the means are not equal, so $H_1:\mu_1-\mu_2
eq0$.

Step2: Determine test statistic

Since the population variances are equal and unknown, and the populations are normally distributed, we use a two - sample t - test. The test statistic is given by $t=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, where $s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}$.

Step3: Calculate pooled standard deviation $s_p$

We have $n_1 = 7$, $\bar{x}_1=44.58$, $s_1 = 7.92$, $n_2=9$, $\bar{x}_2 = 54.40$, $s_2=7.87$. First, calculate $s_p$:
\[

$$\begin{align*} s_p&=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}\\ &=\sqrt{\frac{(7 - 1)\times7.92^2+(9 - 1)\times7.87^2}{7 + 9-2}}\\ &=\sqrt{\frac{6\times62.7264+8\times61.9369}{14}}\\ &=\sqrt{\frac{376.3584 + 495.4952}{14}}\\ &=\sqrt{\frac{871.8536}{14}}\\ &=\sqrt{62.27525714}\\ &\approx7.891 \end{align*}$$

\]

Step4: Calculate test statistic value

\[

$$\begin{align*} t&=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ &=\frac{(44.58 - 54.40)-0}{7.891\sqrt{\frac{1}{7}+\frac{1}{9}}}\\ &=\frac{-9.82}{7.891\sqrt{\frac{9 + 7}{63}}}\\ &=\frac{-9.82}{7.891\sqrt{\frac{16}{63}}}\\ &=\frac{-9.82}{7.891\times0.504}\\ &=\frac{-9.82}{3.977}\\ &\approx - 2.469 \end{align*}$$

\]

Step5: Find critical values

The degrees of freedom is $df=n_1 + n_2-2=7 + 9-2 = 14$. For a two - tailed test with $\alpha = 0.10$, the critical values are $t_{\alpha/2,df}=t_{0.05,14}$. From the t - distribution table, $t_{0.05,14}\approx\pm 1.761$.

Step6: Make a decision

Since $|t|=2.469>1.761$, we reject the null hypothesis.

Answer:

(a) $H_0:\mu_1-\mu_2 = 0$, $H_1:\mu_1-\mu_2
eq0$
(b) Two - sample t - test
(c) $-2.469$
(d) $- 1.761$ and $1.761$
(e) Yes