Answer:
0.000005178
Step-by-step explanation:
we test for the goodness of fit
h0; drink preferences not changed
h1: drink preferences changed
alpha = 0.05
we solve for the test statistics
the chsquare test was performed in the attachment
Chi square = (O-E)²/E
chisquare = 32.30188
degree of freedom = n -1 = 6-1 = 5
using the chi distribution function on excel i calculated the p value
CHIDIST(32.30188, 5) =
P-value = 0.000005178
pvalue is less than 0.05 so we reject null hypothesis and conclude that there is sufficient evidence that drink preferences are changed.
the chisquare calculation is in the attachment. Thank you!
