Question

an electric researcher tested an air fryer listed are prices of the three types of air fryer at alpha 0.10 is there is a significant difference in the average prices of the three types of the air fryer

watt
1000 900 800
270 240 180
245 135 155
190 160 200
215 230 120
250 250 140
230 200 180
200 140
210 130

Answer 1

$$\bar{x_1}=226.25, s_1=27.3535$$

$$\bar{x_2}=185.625, s_2=50.2449$$

$$\bar{x_3}=162.50, s_3=29.6226$$

The following null and alternative hypotheses need to be tested:

$H_0:\sigma_1^2=\sigma_3^2$

$H_1:\sigma_1^2\not=\sigma_3^2$

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

Based on the information provided, the significance level is $\alpha=0.1,$ and the the rejection region for this two-tailed test is $R=\{F:F<0.252\ or\ F>4.876\}.$  

The F-statistic is computed as follows:

$$F= {s_1^2\over s_3^2}={29.6225\over 27.3535}=0.923$$

Since from the sample information we get that $F_L=0.252<0.923=F<4.876=F_R,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma_1^2$ is different than the population variance $\sigma_3^2,$ at the $\alpha=0.1$ significance level.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_3$

$H_1:\mu_1\not=\mu_3$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha=0.1,$ and the degrees of freedom are $df=12.$ In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is $t_c=1.782,$ for $\alpha=0.1$ and$df=12.$

The rejection region for this two-tailed test is $R=\{t:|t|>1.782\}.$ .

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x_1}-\bar{x_3}}{\sqrt{(\dfrac{(n_1-1)s_1^2+(n_3-1)s_3^2}{n_1+n_3-2})(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}=4.168$$

Since it is observed that $|t|=4.168>1.782=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is different than $\mu_3,$ at the 0.1 significance level.