Question

Coffee beans from around the world are used to make coffee. Do coffee tasters prefer coffee beans from particular countries? To investigate this question, a researcher collected coffee aroma ratings based on Arabica coffee beans from two countries, Brazil and Costa Rica. The coffee aroma ratings are summarized in the table below. Each sample contains 10 coffee bean ratings. Sample Standard Deviation n Sample Mean Costa Rica 10 7.74 0.26 Brazil 10 7.54 0.25 We will assume that the aroma ratings of coffees from these two countries are normally distributed. Let's perform a hypothesis test at the 5% significance level. Test that the mean aroma rating of coffee made from Costa Rican coffee beans is greater than the mean aroma rating of coffee made from Brazilian coffee beans. Let be the mean aroma rating of coffee made from Costa Rican coffee beans. And let uz be the mean aroma rating of coffee made from Brazilian coffee beans. The difference in sample means should be: mean for Costa Rican sample minus mean for Brazilian sample. What is the P-value? Use a T-distribution with 18 degrees of freedom. Round to three decimal places.

Answer 1

Ho :  
11 12

Ha :  $\mu_{1}>\mu_{2}$

Level of Significance(l.o.s.) :  $\alpha$ = 0.05

Decision Criteria : Reject Ho at 5% l.o.s. if t cal > t tab,
where t tab = t ($\alpha$, n1+n2-2) = t (0.05, 18) = 1.7341

Calculation : $\bar{x}_{1}$ = 7.74,  $\bar{x}_{2}$ = 7.54, n1 = n2 = 10, s1 = 0.26 & s2 = 0.25

s =   = 7.6407

ni 2 S2 1) * s + (n2 - 1) ni + n2 - 2

t cal =  $\frac{\bar{x}_{1}-\bar{x}_{2}}{s*\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$ = $\frac{7.74-7.54}{7.6407*\sqrt{\frac{1}{10}+\frac{1}{10}}}$ = 0.0585

p-value = P( t > 0.0585 ) = 0.4770

Conclusion : Since t cal < t tab, we fail to reject Ho at 5% l.o.s.& thus, conclude that there is no significant difference between the mean aroma ratings made from Costa Ricon coffee beans & Brazilian coffee beans.

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