Question

In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equal-size towels were used, with four sections of towels tested per brand. The absorbency rating data follow. At a level of significance, does there appear to be a difference in the ability of the brands to absorb water?

Answer 1

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.

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

`H_0:\mu_x=\mu_y=\mu_z`

Alternative hypothesis: the means are not equal

`H_a: \mu_x\\eq\mu_y\\eq\mu_z`

We have to apply a one-way ANOVA

We start by calculating the standard deviation for each brand:

`s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33`

Then, we calculate the mean standard error (MSE):

`MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78`

Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:

`M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91`

The MSB is then:

`s^2=(\sum(M_i-M)^2)/(N-1)\\\\\\s^2=((92-91)^2+(97-91)^2+(84-91)^2)/(3-1)\\\\\\s^2=(1+36+49)/(2)=(86)/(2)=43\\\\\\\\MSB=ns^2=4*43=172`

Now we calculate the F statistic as:

`F=MSB/MSE=172/23.78=7.23`

The degrees of freedom of the numerator are:

`dfn=a-1=3-1=2`

The degrees of freedom of the denominator are:

`dfd=N-a=3*4-3=12-3=9`

The P-value of F=7.23, dfn=2 and dfd=9 is:

`P-value=P(F>7.23)=0.01342`

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.