Halve the significance level and do a one-tailed test.
Eg, two-tailed 10% test of H0: mean = 0 against H1: mean = 0.
If the observed mean is more than 0, do a one-tailed 5% test of H0: mean = 0 against H1: mean > 0.
If the observed mean is less than 0, do a one-tailed 5% test of H0: mean = 0 against H1: mean < 0.
Sorry, could you apply what you just wrote to this question:
In a fruit machine there are 5 drums which rotate independently to show one out of six types of fruit (lemon, apple, orange, melon, banana and pear). You win a prize if all 5 stop showing the same fruit. A customer claims that the machine is fixed; the lemon in the first place is not showing the right number of times. The manager runs the machine 20 times and the lemon shows 6 times in the first place. Is the customer's complaint justified at the 10% significance level?
Sorry, could you apply what you just wrote to this question:
In a fruit machine there are 5 drums which rotate independently to show one out of six types of fruit (lemon, apple, orange, melon, banana and pear). You win a prize if all 5 stop showing the same fruit. A customer claims that the machine is fixed; the lemon in the first place is not showing the right number of times. The manager runs the machine 20 times and the lemon shows 6 times in the first place. Is the customer's complaint justified at the 10% significance level?
X = Number of times the lemon shows in the first place ~ Bin(20, p)
Hypotheses for two-tailed test:
H0: p = 1/6 H1: p = 1/6
Since 6 > 20/6, we look at whether we would reject H0 in a 5% test of H0: p = 1/6 against H1: p > 1/6. Under the null we have P(X >= 6) = 0.1018. Since 0.1018 > 0.05, there is insufficient evidence to reject H0.