Introductory Statistics

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.

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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: \(\hat{p} = \frac{27}{212} = 0.127\text{.}\)

Standard error (or \(SE\)).

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

The standard error is the standard deviation of \(\hat{p}\text{.}\) 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 \(p = 0.1\text{:}\) \(SE = \sqrt{\frac{0.1(1 - 0.1)}{212}} = 0.021\text{.}\) 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: \(SE = \sqrt{\frac{p (1 - p)}{n}} = \sqrt{\frac{0.08 (1 - 0.08)}{800}} = 0.0096\text{.}\)

Standard error.

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

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 \ref{chronic_illness_intro}, 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 \(\hat{p} = \frac{64}{752} = 0.085\text{.}\) The standard error is {\footnotesize[Math: SE &= \sqrt{\frac{p (1 - p)}{n}} \approx \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} &= \sqrt{\frac{0.085 (1 - 0.085)}{752}} = 0.010 ]}%.

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

\(H_0: p = 0.5\) (Neither a majority nor minority of students’ grades improved) \(H_A: p \neq 0.5\) (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.) \(H_A: \mu \neq 15\) (The average amount of company time each employee spends not working is different than 15 minutes for March Madness.).

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.

Set up hypotheses. \(H_0\text{:}\) \(p = 0.5\text{,}\) \(H_A\text{:}\) \(p \neq 0.5\text{.}\) We will use a significance level of \(\alpha = 0.05\text{.}\) (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: \(SE = \sqrt{0.5 (1 - 0.5) / 1000} = 0.016\text{.}\) \(Z = \frac{0.42 - 0.5}{0.016} = -5\text{,}\) 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 \(\alpha = 0.05\text{,}\) 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.).

\(H_0\text{:}\) Anti-depressants do not affect the symptoms of Fibromyalgia. \(H_A\text{:}\) 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.

\(H_0\text{:}\) The restaurant meets food safety and sanitation regulations. \(H_A\text{:}\) 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.

\(H_0: p_{unemp} = p_{underemp}\text{:}\) The proportions of unemployed and underemployed people who are having relationship problems are equal. \(H_A: p_{unemp} \ne p{underemp}\text{:}\) 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.

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: [Math: &H_0: p = 0.5 &&H_A: p \neq 0.5 ] 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. \begin{itemize} \item 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. \item We can compute the sample proportion, \(SE\text{,}\) and test statistic: [Math: \hat{p} &= 19 / 21 = 0.905 SE &= \sqrt{\frac{0.5 \times (1 - 0.5)}{21}} = 0.109 Z &= \frac{0.905 - 0.5}{0.109} = 3.72 ] 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. \item 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 \(H_0\text{,}\) we can conclude that men are being paid higher amounts in ways not explainable by chance alone. \end{itemize} 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: \oiRedirect{textbook-yt_healthcare_triage_gender_pay_gap} {youtu.be/aVhgKSULNQA}.