The key to the hypothesis test problems handed out on Monday April 30:
1. Let µ be the true mean fill. The null hypothesis is H0 : µ = 1.0 versus the alternative
Ha : µ ̸= 1.0. The test statistic is
Z=
x̄ − 1.0
√ .
s/ 50
Plugging in the given data yields Z = −4.01. The area below Z = −4.01 is just
3 × 10−5 , so for the two-tailed test this yields a p-value of 6 × 10−5 , very significant.
We soundly reject H0 at an α = 0.05 significance level.
2. Exactly the same as the last problem, but now the statistic is
t=
x̄ − 1.0
√
s/ 10
which yields t = −1.79. This will have a t-distribution with 9 df. The p-value (twosided) is 0.106 (twice 0.053).
3. The null hypothesis is H0 : µ = 1.0 versus the alternative Ha : µ < 1; (note the actual
null hypothesis that’s implied is µ ≥ 1.0, but we don’t state it this way). Again, the
x̄−1.0
√ . and we’ll reject H0 if Z is too small. Of course we still
test statistic is Z = s/
50
obtain Z = −4.01. The area to the left of Z is 3 × 10−5 , and this is the p-value, very
small. We reject H0 ; the mean fill level is definitely less than 1 gallon.
4. This is a paired comparison, so compute the differences (expenditure minus intake).
You find
Player
1
2
3
4
5
6
7
Difference −0.2 2.9 2.5 2.6 2.5 0.5 1.5
The null hypothesis is H0 : µd = 0 versus Ha : µd ̸= 0. We find d¯ = 1.757, sd = 1.197.
¯
√ should be t distributed with 6 df. The actual value obtained is
The statistic t = sd−0.0
d/ 7
t = 3.884, yielding a p-value of 0.008, pretty significant. We reject H0 .
√
5. Use the statistic Z = (p̂ − p)/ p(1 − p)/n. Under the hypothesis H0 : p = 0.02 we
√
have Z = (p̂ − 0.02)/ (0.02)(0.98)/1000 and with p̂ = 0.028 we obtain Z = 1.807,
significant at the 0.07 level, so we do not reject H0 .
1