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.
