Respuesta :
Answer:
a) There is enough evidence to support the claim that there is a difference in the mean overall distance of brands.
b) The 95% confidence interval for the mean is (0.8, 17.6).
[tex]0.8\leq\mu_1-\mu_2\leq17.6[/tex]
Step-by-step explanation:
The data follow:
Brand 1: 276 279 287 271 283 271 279 275 263 267
Brand 2: 270 238 260 265 273 281 271 270 263 268
Mean and STD for Brand 1
[tex]M_1=\dfrac{1}{10}\sum_{i=1}^{10}(276+279+287+271+283+271+279+275+263+267)\\\\\\ M_1=\dfrac{2751}{10}=275.1[/tex]
[tex]s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\s_1=\sqrt{\dfrac{1}{9}\cdot [(276-(275.1))^2+...+(267-(275.1))^2]}\\\\\\[/tex]
[tex]s_1=\sqrt{\dfrac{1}{9}\cdot [(0.81)+(15.21)+...+(65.61)]}=\sqrt{\dfrac{480.9}{9}}=\sqrt{53.43}\\\\\\s_1=7.31[/tex]
Mean and STD for Brand 2
[tex]M_2=\dfrac{1}{10}\sum_{i=1}^{10}(270+238+260+265+273+281+271+270+263+268)\\\\\\ M_2=\dfrac{2659}{10}=265.9[/tex]
[tex]s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\s_2=\sqrt{\dfrac{1}{9}\cdot [(270-(265.9))^2+(238-(265.9))^2+...+(268-(265.9))^2]}\\\\\\[/tex]
[tex]s_2=\sqrt{\dfrac{1}{9}\cdot [(16.81)+(778.41)+...+(4.41)]}=\sqrt{\dfrac{1164.9}{9}}=\sqrt{129.43}\\\\\\s_2=11.38[/tex]
This is a hypothesis test for the difference between populations means.
The claim is that there is a difference in the mean overall distance of brands.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2> 0[/tex]
The significance level is 0.05.
The sample 1, of size n1=10 has a mean of 275.1 and a standard deviation of 7.31.
The sample 1, of size n1=10 has a mean of 265.9 and a standard deviation of 11.38.
The difference between sample means is Md=9.2.
[tex]M_d=M_1-M_2=275.1-265.9=9.2[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{7.31^2}{10}+\dfrac{11.38^2}{10}}\\\\\\s_{M_d}=\sqrt{5.344+12.95}=\sqrt{18.294}=4.277[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{9.2-0}{4.277}=\dfrac{9.2}{4.277}=2.151[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=10+10-2=18[/tex]
This test is a right-tailed test, with 18 degrees of freedom and t=2.151, so the P-value for this test is calculated as (using a t-table):
[tex]P-value=P(t>2.151)=0.023[/tex]
As the P-value (0.023) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that there is a difference in the mean overall distance of brands.
b) We have to calculate a 95% confidence interval for the difference between means, with a T-model.
The difference between sample means is Md=9.2.
The estimated standard error of the difference is s_Md=4.277.
The critical t-value for a 95% confidence interval is t=1.96.
The margin of error (MOE) can be calculated as:
[tex]MOE=t\cdot s_{M_d}=1.96 \cdot 4.277=8.383[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=M_d-t \cdot s_{M_d} = 9.2-8.383=0.817\\\\UL=M_d+t \cdot s_{M_d} = 9.2+8.383=17.583[/tex]
The 95% confidence interval for the mean is (0.8, 17.6).
[tex]0.8\leq\mu_1-\mu_2\leq17.6[/tex]
