So I have a question with my teacher's solution below and I'm confused about one of the steps in their solution. I would appreciate some clarification.

Two brands of coffee were compared. Two independent random samples of 50 people each were asked to taste either Brand A or Brand B coffee, and indicate whether they liked it or not. Eighty four percent of the people who tasted Brand A liked it; the analogous sample proportion for Brand B was ninety percent.

(A) At α = 0.01, is there a significant difference in the proportions of individuals who like the two coffees? Use the p-value approach.

So I circled in blue a step where I'm confused about in the teacher's solution. I don't understand why P(Z<-0.89) = 0.2061.

On the z-score table -0.89 is equal to 0.1867 so shouldn't it be that value instead? where did 0.2061 come from? Is it possible that the teacher made an error or am I not understanding something?

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  • $\begingroup$ The P-value for a 2-sided test should be $P(|Z| \le 0.89) = P(Z \le -0.89) + P(Z \ge 0.89) = 2P(Z \le -0.89),$ which can be found from printed tables of the standard normal distribution or by using software. In R statistical software, code 2*pnorm(-.89) returns 0.3734659. $\endgroup$
    – BruceET
    Aug 12, 2019 at 8:47

1 Answer 1


Minitab output for this problem is as follows:

Test and CI for Two Proportions 

Sample   X   N  Sample p
1       42  50  0.840000
2       45  50  0.900000

Difference = p (1) - p (2)
Estimate for difference:  -0.06
99% CI for difference:  (-0.232561, 0.112561)
Test for difference = 0 (vs ≠ 0):  
    Z = -0.89  P-Value = 0.372

Notes: (1) The major issue is in doubling the tail probability to get the P-value for a two-sided test. (2) Minitab may be computing the $Z$-statistic to more places than printed. (I think the value to three places is $-0.892).$ If so, that accounts for the small difference in P-values between R and Minitab.


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