Respuesta :
Answer:
Yes. At this significance level, there is evidence to support the claim that there is a difference in the ability of the brands to absorb water.
Step-by-step explanation:
The question is incomplete:
The significance level is 0.05.
The data is:
Brand X: 91, 100, 88, 89
Brand Y: 99, 96, 94, 99
Brand Z: 83, 88, 89, 76
We have to check if there is a significant difference between the absorbency rating of each brand.
Null hypothesis: all means are equal
[tex]H_0:\mu_x=\mu_y=\mu_z[/tex]
Alternative hypothesis: the means are not equal
[tex]H_a: \mu_x\neq\mu_y\neq\mu_z[/tex]
We have to apply a one-way ANOVA
We start by calculating the standard deviation for each brand:
[tex]s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33[/tex]
Then, we calculate the mean standard error (MSE):
[tex]MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78[/tex]
Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:
[tex]M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91[/tex]
The MSB is then:
[tex]s^2=\dfrac{\sum(M_i-M)^2}{N-1}\\\\\\s^2=\dfrac{(92-91)^2+(97-91)^2+(84-91)^2}{3-1}\\\\\\s^2=\dfrac{1+36+49}{2}=\dfrac{86}{2}=43\\\\\\\\MSB=ns^2=4*43=172[/tex]
Now we calculate the F statistic as:
[tex]F=MSB/MSE=172/23.78=7.23[/tex]
The degrees of freedom of the numerator are:
[tex]dfn=a-1=3-1=2[/tex]
The degrees of freedom of the denominator are:
[tex]dfd=N-a=3*4-3=12-3=9[/tex]
The P-value of F=7.23, dfn=2 and dfd=9 is:
[tex]P-value=P(F>7.23)=0.01342[/tex]
As the P-value (0.013) is smaller than the significance level (0.05), the null hypothesis is rejected.
There is evidence to support the claim that there is a difference in the ability of the brands to absorb water.
