Question
1. Which of the following will increase the power of a test?
a. Increase \(\mathrm{n}\).
b. Increase \(\alpha\).
c. Reduce the amount of variability in the sample.
d. Consider an alternative hypothesis further from the null.
e. All of these will increase the power of a test.
▶️Answer/Explanation
Ans:
The correct answer is (e).
Question
Under a null hypothesis, a sample value yields a P-value of 0.015 . Which of the following statements is (are) true?
I. This finding is statistically significant at the 0.05 level of significance.
II. This finding is statistically significant at the 0.01 level of significance.
III. The probability of getting a sample value as extreme as this one by chance alone if the null hypothesis is true is 0.015 .
a. I and III only
b. I only
c. III only
d. II and III only
e. I, II, and III
▶️Answer/Explanation
Ans: The correct answer is (a).
It is not significant at the 0.01 level because 0.015 is greater than 0.01 .
Question
In a test of the null hypothesis \(\mathrm{H}_0: \mathrm{p}=0.35\) with \(\mathrm{a}=0.01\), against the alternative hypothesis \(\mathrm{H}_{\mathrm{A}}: \mathrm{p}<0.35\), a large random sample produced a z-score of -2.05 . Based on this, which of the following conclusions can be drawn?
a. It is likely that \(p<0.35\).
b. \(p<0.35\) only \(2 \%\) of the time.
c. If the z-score were positive instead of negative, we would be able to reject the null hypothesis.
d. We do not have sufficient evidence to claim that \(p<0.35\).
e. \(1 \%\) of the time we will reject the alternative hypothesis in error.
▶️Answer/Explanation
Ans:The correct answer is (d).
To reject the null at the 0.01 level of significance, we would need to have \(z<-2.33\).
Question
A paint manufacturer advertises that one gallon of its paint will cover 400 square feet of interior wall. Some local painters suspect the average coverage is considerably less and decide to conduct an experiment to find out. If \(\mu\) represents the true average number of square feet covered by the paint, which of the following are the correct null and alternative hypotheses to be tested?
a. \(H_0: \mu=400, H_A: \mu>400\)
b. \(\mathrm{H}_0: \mu \geq 400, \mathrm{H}_{\mathrm{A}}: \mu \neq 400\)
c \(\mathrm{H}_0: \mu=400, \mathrm{H}_{\mathrm{A}}: \mu \neq 400\)
d. \(\mathrm{H}_0: \mu \neq 400, \mathrm{H}_{\mathrm{A}}: \mu<400\)
e. \(\mathrm{H}_0: \mu \geq 400, \mathrm{H}_{\mathrm{A}}: \mu<400\)
▶️Answer/Explanation
Ans: The correct answer is (e).
Because we are concerned that the actual amount of coverage might be less than 400 square feet, the only options for the alternative hypothesis are (d) and (e) (the alternative hypothesis in (a) is in the wrong direction, and the alternatives in (b) and (c) are two-sided). The null hypothesis given in (d) is not a form we would use for a null (the only choices are \(=, \leq\), or \(\geq\) ). We might see \(\mathrm{H}_0: \mu=400\) rather than \(\mathrm{H}_0: \mu \geq 400\). Both are correct statements of a null hypothesis against the alternative \(\mathrm{H}_{\mathrm{A}}: \mu<400\).
Question
A school district claims that the average teacher in the district earns \(\$ 48,000\) per year. The teachers’ organization argues that the average salary is less. A random sample of 25 teachers yields a mean salary of \(\$ 47,500\) with a sample standard deviation of \(\$ 2,000\). Assuming that the distribution of all teachers’ salaries is approximately normally distributed, what is the value of the t-test statistic and the P-value for a test of the hypothesis \(\mathrm{H}_0: \mu=48,000\) against \(\mathrm{H}_{\mathrm{A}}: \mu<48,000\) ? (Remember, \(\mathrm{df}=25-1=24\).)
a. \(t=1.25,0.10<P<0.15\)
b. \(t=-1.25,0.20<P<0.30\)
c. \(t=1.25,0.20<P<0.30\)
d. \(t=-1.25,0.10<P<0.15\)
e. \(t=-1.25, P>0.25\)
▶️Answer/Explanation
Ans:The correct answer is (d).
$
t=\frac{47500-48000}{2000 / \sqrt{25}}=-1.25 \Rightarrow 0.10<P<0.15
$
for the one-sided alternative. The calculator answer is \(P=0.112\). Had the alternative been two-sided, the correct answer would have been (b).