- A company produces millions of 1-pound packages of bacon every week. Company specifications allow for no more than 3 percent of the 1-pound packages to be underweight. To investigate compliance with the specifications, the company’s quality control manager selected a random sample of 1,000 packages produced in one week and found 40 packages, or 4 percent, to be underweight. Assuming all conditions for inference are met, do the data provide convincing statistical evidence at the significance level of $\alpha=0.05$ that more than 3 percent of all the packages produced in one week are underweight?
(A) Yes, because the sample estimate of 0.04 is greater than the company specification of 0.03.
(B) Yes, because the p-value of 0.032 is less than the significance level of 0.05.
(C) Yes, because the p-value of 0.064 is greater than the significance level of 0.05.
(D) No, because the p-value of 0.032 is less than the significance level of 0.05.
(E) No, because the p-value of 0.064 is greater than the significance level of 0.05.
The answer is (B) and I was trying to understand why. My calculation was:
$H_0:p=0.03$
$H_a:p>0.03$
$$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$
$$z=\frac{0.04-0.03}{\sqrt{\frac{(0.03)(0.07)}{1000}}}$$
But I get a ridiculously high number. I'm confused about how to get the p-value in this case.
- A two-sided t-test for a population mean is conducted of the null hypothesis $H_0:\mu=100$. If a 90 percent t-interval constructed from the same sample data contains the value of 100, which of the following can be concluded about the test at a significance level of $\alpha=0.10$?
(A) The p-value is less than 0.10, and $H_0$ should be rejected.
(B) The p-value is less than 0.10, and $H_0$ should not be rejected.
(C) The p-value is greater than 0.10, and $H_0$ should be rejected.
(D) The p-value is greater than 0.10, and $H_0$ should not be rejected.
(E) There is not enough information given to make a conclusion about the p-value and $H_0$.
Here the answer is D, but again I am confused. How can I find the p-value in this case?
Best Answer
We can eliminate out of hand the obviously incorrect choices A, C, and D, leaving only B and E as possibly correct.
A is incorrect, because we know that just because the point estimate is above the hypothesized proportion, that does not imply that the variability or uncertainty in that estimate is small enough to state with a high degree of confidence that the true proportion exceeds the hypothesized proportion. Another way to think about it is that if you flip a fair coin $10$ times, there's quite a reasonable chance you may get at least $6$ heads (more than $1$ in $3$ chance), despite the coin being perfectly fair. Yet, if you flip the same coin $1000$ times and get $600$ or more heads, this is extremely improbable (odds less than $1$ in $7$ billion), despite both sample proportions being $0.6 = 6/10 = 600/1000$.
C is incorrect because the conclusion is opposite of what is required by the $p$-value; i.e., if $p > \alpha$, you fail to reject the null hypothesis, whereas the answer choice states that you would reject (Yes) when $0.064 > 0.05$.
D is incorrect for the same reason as C: the conclusion is opposite of what is required. If $p < \alpha$, you would reject the null hypothesis, whereas the answer choice states that you would fail to reject (No) when $0.032 < 0.05$.
This leaves only B and E as choices. Which to choose depends on which $p$-value is correct. Our hypothesis is $$H_0 : p = p_0 = 0.03 \quad \text{vs.} \quad H_a : p > p_0 = 0.03.$$ The test statistic is $$Z \mid H_0 = \frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}} \sim \operatorname{Normal}(0,1).$$ This much you have stated correctly, but the calculation should be $$Z \mid H_0 = \frac{0.04 - 0.03}{\sqrt{0.03 (0.97)/1000}} \approx \frac{0.01}{0.00539444} \approx 1.85376.$$ The $p$-value for this one-sided test is the probability $$\Pr[Z > 1.85376] \approx 0.0318868,$$ which matches the $0.032$ $p$-value in answer choice B.
For your second question, you have insufficient information to compute a $p$-value. What you must do is make a decision based on the information provided. Since the two-sided $90\%$ confidence interval contains the null mean $100$, you know that the two-sided hypothesis test at $\alpha = 1 - 0.9 = 0.10$ must not be able to reject $H_0$ in favor of $H_a$: the uncertainty in your point estimate is too large to rule out with $90\%$ confidence that the true mean is different than $100$, because the CI contains $100$. This also means that $p$-value cannot be less than $\alpha = 0.10$, otherwise you would reject $H_0$. So the correct answer is D.