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The manager of a coffee shop wants to know if his customers’ drink preferences have changed in the past year. He knows that last year the preferences followed the following proportions – 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 =

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shallomisaiah19

Answer:

Step-by-step explanation:

[tex]H_0 : \texttt {null hypothesis}\\\\H_1 : \texttt {alternative hypothesis}[/tex]

The null hypothesis is the drink preferences are not changed at coffee shop.

The alternative hypothesis is the drink preferences are changed at coffee shop.

the level of significance = α = 0.05

We get the Test statistic

[tex]\texttt {Chi square}=\frac{\sum (F_o-F_e)}{F_e}[/tex]

Where, [tex]F_o[/tex] is observed frequencies and

[tex]F_e[/tex] is expected frequencies.

N = 6

Degrees of freedom = df = (N – 1)

= 6 – 1

= 5

the level of significance  α = 0.05

Critical value = 11.07049775

( using Chi square table or excel)

Tables for test statistic are given  below

                             F_o                F_e                 Chi square

Americanos           115                 153                9.4379

Capp.                     88                  94.5             0.447

Espresso               69                  63                 0.5714

Lattes                    59                  49.5               1.823

Macchiatos           44                   45                 0.022

Other                    75                   45                  20

Total                    450                 450                 32.30

[tex]\texttt {Chi square}=\frac{\sum (F_o-F_e)}{F_e}[/tex] = 32.30

P-value = 0.00000517

( using Chi square table or excel)

P-value < α = 0.05

So, we reject the null hypothesis

This is because their sufficient evidence to conclude that Drink preferences are changed at coffee shop.

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