Respuesta :
Answer:
The tabulated value is less than the calculated value, therefore we accept the null hypothesis and It show that there is a difference in the mean overall distance of brands.
Step-by-step explanation:
The given data sets are
Brand 1: 277 278 287 271 283 271 279 275 263 267
Brand 2: 262 248 260 265 273 281 271 270 263
We need to check that whether there is a difference in the mean overall distance of brands or not.
[tex]\overline{x}_1=275.1[/tex]
[tex]\overline{x}_2=265.89[/tex]
[tex]s_1=7.2793[/tex]
[tex]s_2=9.3601[/tex]
Null hypothesis:
[tex]H_0:\mu_1=\mu_2[/tex]
Alternative hypothesis:
[tex]H_a:\mu_1\neq \mu_2[/tex]
The data is normally distributed, we assume that the variances are equal so we will apply t-test.
The formula for t-statistics is
[tex]t=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{s\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex]
where,
[tex]s=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}[/tex]
[tex]s=\sqrt{\frac{(9)(7.2793)^2+(9)(9.3601)^2}{10+10-2}}[/tex]
[tex]s=8.3845[/tex]
The value of t is
[tex]t=\frac{(265.89-275.1)-(0)}{8.3845\sqrt{\frac{1}{10}+\frac{1}{10}}}[/tex]
[tex]t=-2.456[/tex]
[tex]|t|=2.456[/tex]
Degree of freedom is
[tex]d.f.=10+10-2[/tex]
From t-table the t-value for 0.05 level of significance at 18 degree of freedom is ±2.1009.
Since the tabulated value is less than the calculated value, therefore we accept the null hypothesis and It show that there is a difference in the mean overall distance of brands.
