Chapter 5: Foundations for inferenceContent check — Statistics (OpenIntro), Chapter 5: Foundations for inference

Since the sample size $n$ is in the denominator (on the bottom) of the fraction, a bigger sample size means the entire expression when calculated will tend to be smaller. That is, a larger sample size would correspond to a smaller standard error.

Mean. Each student reports a numerical value: a number of hours.

Mean. Each student reports a number, which is a percentage, and we can average over these percentages.

Proportion. Each student reports Yes or No, so this is a categorical variable and we use a proportion.

Mean. Each student reports a number, which is a percentage like in part (b).

Proportion. Each student reports whether or not s/he expects to get a job, so this is a categorical variable and we use a proportion.

The sample is from all computer chips manufactured at the factory during the week of production. We might be tempted to generalize the population to represent all weeks, but we should exercise caution here since the rate of defects may change over time.

The fraction of computer chips manufactured at the factory during the week of production that had defects.

Estimate the parameter using the data: $\hat{p}=\frac{27}{212}=0.127$⁠.

Standard error (or $SE$⁠).

Compute the $SE$ using $\hat{p}=0.127$ in place of $p$⁠: $SE\approx$ $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=$ $\sqrt{\frac{0.127(1-0.127)}{212}}=$ $0.023$⁠.

The standard error is the standard deviation of $\hat{p}$⁠. A value of 0.10 would be about one standard error away from the observed value, which would not represent a very uncommon deviation. (Usually beyond about 2 standard errors is a good rule of thumb.) The engineer should not be surprised.

Recomputed standard error using $p=0.1$⁠: $SE=\sqrt{\frac{0.1(1-0.1)}{212}}=0.021$⁠. This value isn’t very different, which is typical when the standard error is computed using relatively similar proportions (and even sometimes when those proportions are quite different!).

Sampling distribution.

If the population proportion is in the 5-30% range, the success-failure condition would be satisfied and the sampling distribution would be symmetric.

We use the formula for the standard error: $SE=$ $\sqrt{\frac{p(1-p)}{n}}=$ $\sqrt{\frac{0.08(1-0.08)}{800}}=$ $0.0096$⁠.

Standard error.

The distribution will tend to be more variable when we have fewer observations per sample.

If we want to be more certain we will capture the fish, we might use a wider net. Likewise, we use a wider confidence interval if we want to be more certain that we capture the parameter.

This is equivalent to asking how often the Z-score will be larger than -2.58 but less than 2.58. For a picture, see Figure 5.7. To determine this probability, we can use statistical software, a calculator, or a table to look up -2.58 and 2.58 for a normal distribution: 0.0049 and 0.9951. Thus, there is a $0.9951-0.0049\approx 0.99$ probability that an unobserved normal random variable $X$ will be within 2.58 standard deviations of $\mu$⁠.

If we took many such samples and computed a 95% confidence interval for each, then about 95% of those intervals would contain the actual proportion of New York adults who supported a quarantine for anyone who has come into contact with an Ebola patient.

Not necessarily. The poll was taken at a time where there was a huge public safety concern. Now that people have had some time to step back, they may have changed their opinions. We would need to run a new poll if we wanted to get an estimate of the current proportion of New York adults who would support such a quarantine period.

The survey was a random sample and counts are both $\ge 10$ ($1000\times 0.848=848$ and $1000\times 0.152=152$⁠), so independence and the success-failure condition are satisfied, and $\hat{p}=0.848$ can be modeled using a normal distribution.

Guided Practice 5.15 confirmed that $\hat{p}$ closely follows a normal distribution, so we can use the C.I. formula:

$$\begin{array}{c}\hfill \text{point estimate}\pm z^{\star }\times SE\end{array}$$

In this case, the point estimate is $\hat{p}=0.848$⁠. For a 99% confidence interval, $z^{\star }=2.58$⁠. Computing the standard error: $S{E}_{\hat{p}}=$ $\sqrt{\frac{0.848(1-0.848)}{1000}}=$ $0.0114$⁠. Finally, we compute the interval as $0.848\pm 2.58\times 0.0114\to$ $(0.8186,0.8774)$⁠. It is also important to always provide an interpretation for the interval: we are 99% confident the proportion of American adults that support expanding the use of wind turbines in 2018 is between 81.9% and 87.7%.

No, a confidence interval only provides a range of plausible values for a parameter, not future point estimates.

Recall that the general formula is $pointestimate\pm z^{\star }\times SE$⁠. First, identify the three different values. The point estimate is 45%, $z^{\star }=1.96$ for a 95% confidence level, and $SE=1.2\%$⁠. Then, plug the values into the formula: $45\%\pm 1.96\times 1.2\%$ $\to$ $(42.6\%,47.4\%)$ We are 95% confident that the proportion of US adults who live with one or more chronic conditions is between 42.6% and 47.4%.

False. Confidence intervals provide a range of plausible values, and sometimes the truth is missed. A 95% confidence interval “misses” about 5% of the time.

True. Notice that the description focuses on the true population value.

True. If we examine the 95% confidence interval computed in Exercise 5.7, we can see that 50% is not included in this interval. This means that in a hypothesis test, we would reject the null hypothesis that the proportion is 0.5.

False. The standard error describes the uncertainty in the overall estimate from natural fluctuations due to randomness, not the uncertainty corresponding to individuals’ responses.

False. The point estimate is always in the confidence interval, and this is a non-sensical use of a confidence interval with a point estimate (because the point estimate is, by design, listed within the confidence interval).

True.

False. The confidence interval is not about a sample mean.

False. To be more confident that we capture the parameter, we need a wider interval. Think about needing a bigger net to be more sure of catching a fish in a murky lake.

True. Optional explanation: This is true since the normal model was used to model the sample mean. The margin of error is half the width of the interval, and the sample mean is the midpoint of the interval.

False. In the calculation of the standard error, we divide the standard deviation by the square root of the sample size. To cut the SE (or margin of error) in half, we would need to sample $2^2=4$ times the number of people in the initial sample.

The visitors are from a simple random sample, so independence is satisfied. The success-failure condition is also satisfied, with both 64 and $752-64=688$ above 10. Therefore, we can use a normal distribution to model $\hat{p}$ and construct a confidence interval.

The sample proportion is $\hat{p}=\frac{64}{752}=0.085$⁠. The standard error is

$$\begin{array}{cc}\hfill SE& =\sqrt{\frac{p(1-p)}{n}}\approx \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\hfill \\ \hfill & =\sqrt{\frac{0.085(1-0.085)}{752}}=0.010\hfill \end{array}$$

For a 90% confidence interval, use $z^{\star }=1.6449$⁠. The confidence interval is $0.085\pm 1.6449\times 0.010\to$ $(0.0683,0.1017)$⁠. We are 90% confident that 6.83% to 10.17% of first-time site visitors will register using the new design.

The jury considers whether the evidence is so convincing (strong) that there is no reasonable doubt regarding the person’s guilt; in such a case, the jury rejects innocence (the null hypothesis) and concludes the defendant is guilty (alternative hypothesis).

The null hypothesis ($H_0$⁠) in this case is the declaration of no difference: the drugs are equally effective. The alternative hypothesis ($H_A$⁠) is that the new drug performs differently than the original, i.e. it could perform better or worse.

The appropriate hypotheses are:

$H_0$⁠: the proportion who get the answer correct is the same as random guessing: 1-in-3, or $p=0.333$⁠.

$H_A$⁠: the proportion who get the answer correct is different than random guessing, $p\ne 0.333$⁠.

The correct answer to the question is 2 billion. While the world population is projected to increase, the average age is also expected to rise. That is, the majority of the population growth will happen in older age groups, meaning people are projected to live longer in the future across much of the world.

We check both conditions, which are satisfied, so it is reasonable to use a normal distribution for $\hat{p}$⁠:
Independence. Since the data are from a simple random sample, the observations are independent.
Success-failure. We’ll use $\hat{p}$ in place of $p$ to check: $n\hat{p}=34$ and $n(1-\hat{p})=194$⁠. Both are greater than 10, so the success-failure condition is satisfied.

If the court makes a Type 1 Error, this means the defendant is innocent ($H_0$ true) but wrongly convicted. Note that a Type 1 Error is only possible if we’ve rejected the null hypothesis.

A Type 2 Error means the court failed to reject $H_0$ (i.e. failed to convict the person) when she was in fact guilty ($H_A$ true). Note that a Type 2 Error is only possible if we have failed to reject the null hypothesis.

To lower the Type 2 Error rate, we want to convict more guilty people. We could lower the standards for conviction from “beyond a reasonable doubt” to “beyond a little doubt”. Lowering the bar for guilt will also result in more wrongful convictions, raising the Type 1 Error rate.

We would like to understand if a majority supports or opposes, or ultimately, if there is no difference. If $p$ is the proportion of Americans who support nuclear arms reduction, then $H_0$⁠: $p=0.50$ and $H_A$⁠: $p\ne 0.50$⁠.

Here the null hypothesis is that the part is not broken, and the alternative is that it is broken. If we don’t have sufficient evidence to reject $H_0$⁠, we would not replace the part. It sounds like failing to fix the part if it is broken ($H_0$ false, $H_A$ true) is not very problematic, and replacing the part is expensive. Thus, we should require very strong evidence against $H_0$ before we replace the part. Choose a small significance level, such as $\alpha =0.01$⁠.

$H_0:p=0.5$ (Neither a majority nor minority of students’ grades improved) $H_A:p\ne 0.5$ (Either a majority or a minority of students’ grades improved)

$H_0:\mu =15$ (The average amount of company time each employee spends not working is 15 minutes for March Madness.) $H_A:\mu \ne 15$ (The average amount of company time each employee spends not working is different than 15 minutes for March Madness.)

(1) The hypotheses should be about the population proportion ($p$⁠), not the sample proportion. (2) The null hypothesis should have an equal sign. (3) The alternative hypothesis should have a not-equals sign, and (4) it should reference the null value, $p_0=0.6$⁠, not the observed sample proportion. The correct way to set up these hypotheses is: $H_0:p=0.6$ and $H_A:p\ne 0.6$⁠.

This claim is reasonable, since the entire interval lies above 50%.

The value of 70% lies outside of the interval, so we have convincing evidence that the researcher’s conjecture is wrong.

A 90% confidence interval will be narrower than a 95% confidence interval. Even without calculating the interval, we can tell that 70% would not fall in the interval, and we would reject the researcher’s conjecture based on a 90% confidence level as well.

(i) Set up hypotheses. $H_0$⁠: $p=0.5$⁠, $H_A$⁠: $p\ne 0.5$⁠. We will use a significance level of $\alpha =0.05$⁠. (ii) Check conditions: simple random sample gets us independence, and the success-failure conditions is satisfied since $0.5\times 1000=500$ for each group is at least 10. (iii) Next, we calculate: $SE=$ $\sqrt{0.5(1-0.5)/1000}=$ $0.016$⁠. $Z=\frac{0.42-0.5}{0.016}=-5$⁠, which has a one-tail area of about 0.0000003, so the p-value is twice this one-tail area at 0.0000006. (iv) Make a conclusion: Because the p-value is less than $\alpha =0.05$⁠, we reject the null hypothesis and conclude that the fraction of US adults who believe raising the minimum wage will help the economy is not 50%. Because the observed value is less than 50% and we have rejected the null hypothesis, we can conclude that this belief is held by fewer than 50% of US adults. (For reference, the survey also explores support for changing the minimum wage, which is a different question than if it will help the economy.)

If the p-value is 0.05, this means the test statistic would be either $Z=-1.96$ or $Z=1.96$⁠. We’ll show the calculations for $Z=1.96$⁠. Standard error: $SE=\sqrt{0.3(1-0.3)/90}=0.048$⁠. Finally, set up the test statistic formula and solve for $\hat{p}$⁠: $1.96=$ $\frac{\hat{p}-0.3}{0.048}\to$ $\hat{p}=$ $0.394$ Alternatively, if $Z=-1.96$ was used: $\hat{p}=0.206$⁠.

$H_0$⁠: Anti-depressants do not affect the symptoms of Fibromyalgia. $H_A$⁠: Anti-depressants do affect the symptoms of Fibromyalgia (either helping or harming).

Concluding that anti-depressants either help or worsen Fibromyalgia symptoms when they actually do neither.

Concluding that anti-depressants do not affect Fibromyalgia symptoms when they actually do.

We are 95% confident that Americans spend an average of 1.38 to 1.92 hours per day relaxing or pursuing activities they enjoy.

Their confidence level must be higher as the width of the confidence interval increases as the confidence level increases.

The new margin of error will be smaller, since as the sample size increases, the standard error decreases, which will decrease the margin of error.

$H_0$⁠: The restaurant meets food safety and sanitation regulations. $H_A$⁠: The restaurant does not meet food safety and sanitation regulations.

The food safety inspector concludes that the restaurant does not meet food safety and sanitation regulations and shuts down the restaurant when the restaurant is actually safe.

The food safety inspector concludes that the restaurant meets food safety and sanitation regulations and the restaurant stays open when the restaurant is actually not safe.

A Type 1 Error may be more problematic for the restaurant owner since his restaurant gets shut down even though it meets the food safety and sanitation regulations.

A Type 2 Error may be more problematic for diners since the restaurant deemed safe by the inspector is actually not.

Strong evidence. Diners would rather a restaurant that meet the regulations get shut down than a restaurant that doesn’t meet the regulations not get shut down.

$H_0:p_{unemp}=p_{underemp}$⁠: The proportions of unemployed and underemployed people who are having relationship problems are equal. $H_A:p_{unemp}\ne punderemp$⁠: The proportions of unemployed and underemployed people who are having relationship problems are different.

If in fact the two population proportions are equal, the probability of observing at least a 2% difference between the sample proportions is approximately 0.35. Since this is a high probability we fail to reject the null hypothesis. The data do not provide convincing evidence that the proportion of of unemployed and underemployed people who are having relationship problems are different.

Because 130 is inside the confidence interval, we do not have convincing evidence that the true average is any different than what the nutrition label suggests.

True. If the sample size gets ever larger, then the standard error will become ever smaller. Eventually, when the sample size is large enough and the standard error is tiny, we can find statistically significant yet very small differences between the null value and point estimate (assuming they are not exactly equal).

In effect, we’re checking whether men are paid more than women (or vice-versa), and we’d expect these outcomes with either chance under the null hypothesis:

$$\begin{array}{cccc}\hfill & H_0:p=0.5\hfill & \hfill & H_A:p\ne 0.5\hfill \end{array}$$

We’ll use $p$ to represent the fraction of cases where men are paid more than women.

Below is the completion of the hypothesis test.

• There isn’t a good way to check independence here since the jobs are not a simple random sample. However, independence doesn’t seem unreasonable, since the individuals in each job are different from each other. The success-failure condition is met since we check it using the null proportion: $p_{0}n=(1-p_0)n=10.5$ is greater than 10.

• We can compute the sample proportion, $SE$⁠, and test statistic:
$$\begin{array}{cc}\hfill \hat{p}& =19/21=0.905\hfill \\ \hfill SE& =\sqrt{\frac{0.5\times (1-0.5)}{21}}=0.109\hfill \\ \hfill Z& =\frac{0.905-0.5}{0.109}=3.72\hfill \end{array}$$
The test statistic $Z$ corresponds to an upper tail area of about 0.0001, so the p-value is 2 times this value: 0.0002.

• Because the p-value is smaller than 0.05, we reject the notion that all these gender pay disparities are due to chance. Because we observe that men are paid more in a higher proportion of cases and we have rejected $H_0$⁠, we can conclude that men are being paid higher amounts in ways not explainable by chance alone.

If you’re curious for more info around this topic, including a discussion about adjusting for additional factors that affect pay, please see the following video by Healthcare Triage: youtu.be/aVhgKSULNQA.