cheergirl2133
02.07.2020 •
Mathematics
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. AmericanosCapp. Observed Counts1158869594475 Expected Counts15394.56349.54545 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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Ответ:
Step-by-step explanation:
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
Where, is observed frequencies and
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
= 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.
Ответ:
This means that:
w<1 kg