Question

A schoolteacher would like to know whether or not toothpaste brands are used differentially in her classroom (in other words, is one brand preferred over the others?). She asks her students to report which of three brands they use: Crest, Colgate, or Aquafresh. Below are the numbers of students who use each type of toothpaste (notice there are 45 students total in her class). Test her hypothesis using an alpha level of .05.
Crest Colgate Aquafresh
24 13 8
a. What test is appropriate for this analysis?
b. State the null hypothesis:
c. State the alternative hypothesis:
d. Find the critical value:
e. Calculate the test statistic:
f. Make a decision:

Answer 1

Answer:

Step-by-step explanation:

a. What test is appropriate for this analysis?

A Chi-square  test of independence is appropriate for this analysis because it is used to compare two variables or testing relationship on categorical variables.  

b. State the null hypothesis:

The null hypothesis is the default hypothesis

$\mathtt{H_o:}$ There is no particular preference for any brand of toothpaste among students.

c. State the alternative hypothesis:

The alternative hypothesis is the research hypothesis which comes in place to challenge the validity of the null hypothesis.

$\mathtt{H_a:}$ There is particular preference for brands of toothpaste among students.

d. Find the critical value:

degree of freedom = n-1

degree of freedom = 3 - 1

degree of freedom = 2

At the level of significance ∝ = 0.05

The confidence interval = 0.95 and degree of freedom = 2, the critical value from the chi-square distribution table = 5.991

e. Calculate the test statistic:

Using the chi square test statistics; we have the following:

Crest             Colgate            Aquafresh            Total

24                        13                     8                      45

Since we have three brands. Then, for each brand, the expected value

= Total /3

= 45/3

=15

Thus:

Chi -square $\mathtt{X^2 = \dfrac{(observed \ value - expected \ value)^2}{expected \ value}}$

$\mathtt{X^2 = \dfrac{(24 - 15)^2}{15} + \dfrac{(13 - 15)^2}{15} + \dfrac{(8 - 15)^2}{15} }$

$\mathtt{X^2 = \dfrac{81}{15} + \dfrac{4}{15} + \dfrac{49}{15} }$

$\mathtt{X^2 = \dfrac{81+4+49}{15}}$

$\mathtt{X^2 = \dfrac{134}{15}}$

$\mathtt{X^2 =8.93 }$

f. Make a decision:

Since the chi-square value is greater than the critical value , we reject the null hypothesis and conclude that the students have  particular preference for brands of toothpaste.