s – 34% Americano, 21% Cappuccino, 14% Espresso, 11% Latte, 10% Macchiato, 10% Other. In a random sample of 450 customers, he finds that 115 ordered Americanos, 88 ordered Cappuccinos, 69 ordered Espressos, 59 ordered Lattes, 44 ordered Macchiatos, and the rest ordered something in the Other category. Run a Goodness of Fit test to determine whether or not drink preferences have changed at his coffee shop. Use a 0.05 level of significance. Americanos Capp. Espresso Lattes Macchiatos Other Observed Counts 115 88 69 59 44 75 Expected Counts 153 94.5 63 49.5 45 45 Enter the p-value - round to 5 decimal places. Make sure you put a 0 in front of the decimal. P-value =___

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!

answer: check all that apply, where the options?

step-by-step explanation:

c

step-by-step explanation: