Chapter 6: Inference for categorical dataContent check — Statistics (OpenIntro), Chapter 6: Inference for categorical data

$SE=$ $\sqrt{\frac{p(1-p)}{n}}\approx$ $\sqrt{\frac{0.70(1-0.70)}{826}}=$ $0.016$⁠.

$H_0$⁠: $p=0.50$⁠. $H_A$⁠: $p\ne 0.50$⁠.

Independence holds since the poll is based on a random sample. The success-failure condition also holds, which is checked using the null value ($p_0=0.5$⁠) from $H_0$⁠: $n{p}_0=826\times 0.5=413$⁠, $n(1-p_0)=826\times 0.5=413$⁠.

For a 90% confidence interval, $z^{\star }=1.6449$⁠, and since an estimate of the proportion 0.017 is available, we’ll use it in the margin of error formula:

$$\begin{array}{cc}\hfill 1.6449\times \sqrt{\frac{0.017(1-0.017)}{n}}& <0.01\to \frac{0.017(1-0.017)}{n}<{\left(\frac{0.01}{1.6449}\right)}^2\to 452.15<n\hfill \end{array}$$

For sample size calculations, we always round up, so the first tire model suggests 453 tires would be sufficient.

A similar computation can be accomplished using 0.062 and 0.013 for $p$⁠, and you should verify that using these proportions results in minimum sample sizes of 1574 and 348 tires, respectively.

We complete the same computations as before, except now we use $0.70$ instead of $0.5$ for $p$⁠:

$$\begin{array}{cc}\hfill 1.96\times \sqrt{\frac{p(1-p)}{n}}\approx 1.96\times \sqrt{\frac{0.70(1-0.70)}{n}}& \le 0.05\to n\ge 322.7\hfill \end{array}$$

A sample size of 323 or more would be reasonable. (Reminder: always round up for sample size calculations!) Given that we plan to track this poll over time, we also may want to periodically repeat these calculations to ensure that we’re being thoughtful in our sample size recommendations in case the baseline rate fluctuates.

False. Doesn’t satisfy success-failure condition.

True. The success-failure condition is not satisfied. In most samples we would expect $\hat{p}$ to be close to 0.08, the true population proportion. While $\hat{p}$ can be much above 0.08, it is bound below by 0, suggesting it would take on a right skewed shape. Plotting the sampling distribution would confirm this suspicion.

False. $S{E}_{\hat{p}}=0.0243$⁠, and $\hat{p}=0.12$ is only $\frac{0.12-0.08}{0.0243}=1.65$ SEs away from the mean, which would not be considered unusual.

True. $\hat{p}=0.12$ is 2.32 standard errors away from the mean, which is often considered unusual.

False. Decreases the SE by a factor of $1/\sqrt{2}$⁠.

True. See the reasoning of 6.1(b).

True. We take the square root of the sample size in the SE formula.

True. The independence and success-failure conditions are satisfied.

True. The independence and success-failure conditions are satisfied.

False. A confidence interval is constructed to estimate the population proportion, not the sample proportion.

True. 95% CI: $82\%\pm 2\%$⁠.

True. By the definition of the confidence level.

True. Quadrupling the sample size decreases the SE and ME by a factor of $1/\sqrt{4}$⁠.

True. The 95% CI is entirely above 50%.

With a random sample, independence is satisfied. The success-failure condition is also satisfied. $ME=$ $z^{\star }\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=$ $1.96\sqrt{\frac{0.56\times 0.44}{600}}=$ $0.0397\approx$ $4\%$

No. The sample only represents students who took the SAT, and this was also an online survey.

(0.5289, 0.5711). We are 90% confident that 53% to 57% of high school seniors who took the SAT are fairly certain that they will participate in a study abroad program in college.

90% of such random samples would produce a 90% confidence interval that includes the true proportion.

Yes. The interval lies entirely above 50%.

We want to check for a majority (or minority), so we use the following hypotheses:

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

We have a sample proportion of $\hat{p}=0.55$ and a sample size of $n=617$ independents.
Since this is a random sample, independence is satisfied. The success-failure condition is also satisfied: $617\times 0.5$ and $617\times (1-0.5)$ are both at least 10 (we use the null proportion $p_0=0.5$ for this check in a one-proportion hypothesis test).
Therefore, we can model $\hat{p}$ using a normal distribution with a standard error of

$$\begin{array}{c}\hfill SE=\sqrt{\frac{p(1-p)}{n}}=0.02\end{array}$$

(We use the null proportion $p_0=0.5$ to compute the standard error for a one-proportion hypothesis test.) Next, we compute the test statistic:

$$\begin{array}{c}\hfill Z=\frac{0.55-0.5}{0.02}=2.5\end{array}$$

This yields a one-tail area of 0.0062, and a p-value of $2\times 0.0062=0.0124$⁠.
Because the p-value is smaller than 0.05, we reject the null hypothesis. We have strong evidence that the support is different from 0.5, and since the data provide a point estimate above 0.5, we have strong evidence to support this claim by the TV pundit.

No. Generally we expect a hypothesis test and a confidence interval to align, so we would expect the confidence interval to show a range of plausible values entirely above 0.5. However, if the confidence level is misaligned (e.g. a 99% confidence level and a $\alpha =0.05$ significance level), then this is no longer generally true.

$H_0:p=0.5$⁠. $H_A:p\ne 0.5$⁠. Independence (random sample) is satisfied, as is the success-failure conditions (using $p_0=0.5$⁠, we expect 40 successes and 40 failures). $Z=2.91$ $\to$ the one tail area is 0.0018, so the p-value is 0.0036. Since the p-value $<0.05$⁠, we reject the null hypothesis. Since we rejected $H_0$ and the point estimate suggests people are better than random guessing, we can conclude the rate of correctly identifying a soda for these people is significantly better than just by random guessing.

If in fact people cannot tell the difference between diet and regular soda and they were randomly guessing, the probability of getting a random sample of 80 people where 53 or more identify a soda correctly (or 53 or more identify a soda incorrectly) would be 0.0036.

Because a sample proportion ($\hat{p}=0.55$⁠) is available, we use this for the sample size calculations. The margin of error for a 90% confidence interval is $1.6449\times SE=$ $1.6449\times \sqrt{\frac{p(1-p)}{n}}$⁠. We want this to be less than 0.01, where we use $\hat{p}$ in place of $p$⁠:

$$\begin{array}{c}\hfill 1.6449\times \sqrt{\frac{0.55(1-0.55)}{n}}\le 0.01\\ \hfill {1.6449}^2\frac{0.55(1-0.55)}{{0.01}^2}\le n\end{array}$$

From this, we get that $n$ must be at least 6697.

Because the patients were randomized, the subjects are independent, both within and between the two groups. The success-failure condition is also met for both groups as all counts are at least 10. This satisfies the conditions necessary to model the difference in proportions using a normal distribution.

Compute the sample proportions (${\hat{p}}_{\text{fish oil}}=0.0112$⁠, ${\hat{p}}_{\text{placebo}}=0.0155$⁠), point estimate of the difference ($0.0112-0.0155=-0.0043$⁠), and standard error ($SE=$ $\sqrt{\frac{0.0112\times 0.9888}{12933}+\frac{0.0155\times 0.9845}{12938}}=$ $0.00145$⁠). Next, plug the values into the general formula for a confidence interval, where we’ll use a 95% confidence level with $z^{\star }=1.96$⁠:

$$\begin{array}{c}\hfill -0.0043\pm 1.96\times 0.00145\to (-0.0071,-0.0015)\end{array}$$

We are 95% confident that fish oils decreases heart attacks by 0.15 to 0.71 percentage points (off of a baseline of about 1.55%) over a 5-year period for subjects who are similar to those in the study. Because the interval is entirely below 0, the data provide strong evidence that fish oil supplements reduce heart attacks in patients like those in the study.

This is an experiment. Patients were randomized to receive mammograms or a standard breast cancer exam. We will be able to make causal conclusions based on this study.

$H_0$⁠: the breast cancer death rate for patients screened using mammograms is the same as the breast cancer death rate for patients in the control, $p_{mgm}-p_{ctrl}=0$⁠.
$H_A$⁠: the breast cancer death rate for patients screened using mammograms is different than the breast cancer death rate for patients in the control, $p_{mgm}-p_{ctrl}\ne 0$⁠.

$H_0$⁠: The higher-quality blades will pass inspection 3% more frequently than the standard-quality blades. $p_{highQ}-p_{standard}=0.03$⁠. $H_A$⁠: The higher-quality blades will pass inspection some amount different than 3% more often than the standard-quality blades. $p_{highQ}-p_{standard}\ne 0.03$⁠.

The standard error squared represents the variance of the estimate. If $X$ and $Y$ are two random variables with variances ${\sigma }_x^2$ and ${\sigma }_y^2$⁠, then the variance of $X-Y$ is ${\sigma }_x^2+{\sigma }_y^2$⁠. Likewise, the variance corresponding to ${\hat{p}}_1-{\hat{p}}_2$ is ${\sigma }_{{\hat{p}}_1}^2+{\sigma }_{{\hat{p}}_2}^2$⁠. Because ${\sigma }_{{\hat{p}}_1}^2$ and ${\sigma }_{{\hat{p}}_2}^2$ are just another way of writing $S{E}_{{\hat{p}}_1}^2$ and $S{E}_{{\hat{p}}_2}^2$⁠, the variance associated with ${\hat{p}}_1-{\hat{p}}_2$ may be written as $S{E}_{{\hat{p}}_1}^2+S{E}_{{\hat{p}}_2}^2$⁠.

This is not a randomized experiment, and it is unclear whether people would be affected by the behavior of their peers. That is, independence may not hold. Additionally, there are only 5 interventions under the provocative scenario, so the success-failure condition does not hold. Even if we consider a hypothesis test where we pool the proportions, the success-failure condition will not be satisfied. Since one condition is questionable and the other is not satisfied, the difference in sample proportions will not follow a nearly normal distribution.

False. The entire confidence interval is above 0.

True.

True.

True.

False. It is simply the negated and reordered values: (-0.06,-0.02).

Standard error:

$$\begin{array}{c}\hfill SE=\sqrt{\frac{0.79(1-0.79)}{347}+\frac{0.55(1-0.55)}{617}}=0.03\end{array}$$

Using $z^{\star }=1.96$⁠, we get:

$$\begin{array}{c}\hfill 0.79-0.55\pm 1.96\times 0.03\to (0.181,0.299)\end{array}$$

We are 95% confident that the proportion of Democrats who support the plan is 18.1% to 29.9% higher than the proportion of Independents who support the plan.

True.

College grads: 23.7%. Non-college grads: 33.7%.

Let $p_{CG}$ and $p_{NCG}$ represent the proportion of college graduates and non-college graduates who responded “do not know". $H_0:p_{CG}=p_{NCG}$⁠. $H_A:p_{CG}\ne p_{NCG}$⁠. Independence is satisfied (random sample), and the success-failure condition, which we would check using the pooled proportion (${\hat{p}}_{\text{𝑝𝑜𝑜𝑙}}=$ $235/827=$ $0.284$⁠), is also satisfied. $Z=-3.18$ $\to$ p-value = 0.0014. Since the p-value is very small, we reject $H_0$⁠. The data provide strong evidence that the proportion of college graduates who do not have an opinion on this issue is different than that of non-college graduates. The data also indicate that fewer college grads say they “do not know” than non-college grads (i.e. the data indicate the direction after we reject $H_0$⁠).

College grads: 35.2%. Non-college grads: 33.9%.

Let $p_{CG}$ and $p_{NCG}$ represent the proportion of college graduates and non-college grads who support offshore drilling. $H_0:p_{CG}=p_{NCG}$⁠. $H_A:p_{CG}\ne p_{NCG}$⁠. Independence is satisfied (random sample), and the success-failure condition, which we would check using the pooled proportion (${\hat{p}}_{\text{𝑝𝑜𝑜𝑙}}=$ $286/827=$ $0.346$⁠), is also satisfied. $Z=0.39$ $\to$ p-value $=0.6966$⁠. Since the p-value $>\alpha$ (0.05), we fail to reject $H_0$⁠. The data do not provide strong evidence of a difference between the proportions of college graduates and non-college graduates who support off-shore drilling in California.

Subscript ${}_C$ means control group. Subscript ${}_T$ means truck drivers. $H_0:p_C=p_T$⁠. $H_A:p_C\ne p_T$⁠. Independence is satisfied (random samples), as is the success-failure condition, which we would check using the pooled proportion (${\hat{p}}_{\text{𝑝𝑜𝑜𝑙}}=70/495=0.141$⁠). $Z=-1.65$ $\to$ p-value $=0.0989$⁠. Since the p-value is high (default to alpha = 0.05), we fail to reject $H_0$⁠. The data do not provide strong evidence that the rates of sleep deprivation are different for non-transportation workers and truck drivers.

Summary of the study:

$H_0:p_N=p_L$⁠. There is no difference in virologic failure rates between the Nevaripine and Lopinavir groups. $H_A:p_N\ne p_L$⁠. There is some difference in virologic failure rates between the Nevaripine and Lopinavir groups.

Random assignment was used, so the observations in each group are independent. If the patients in the study are representative of those in the general population (something impossible to check with the given information), then we can also confidently generalize the findings to the population. The success-failure condition, which we would check using the pooled proportion (${\hat{p}}_{pool}=36/240=0.15$⁠), is satisfied. $Z=2.89$ $\to$ p-value $=0.0039$⁠. Since the p-value is low, we reject $H_0$⁠. There is strong evidence of a difference in virologic failure rates between the Nevaripine and Lopinavir groups. Treatment and virologic failure do not appear to be independent.

The center becomes larger. If took a careful look, we could see that the mean of each distribution is equal to the distribution’s degrees of freedom.

The variability increases as the degrees of freedom increases.

The distribution is very strongly skewed for $df=2$⁠, and then the distributions become more symmetric for the larger degrees of freedom $df=4$ and $df=9$⁠. We would see this trend continue if we examined distributions with even more larger degrees of freedom.

The area is 0.1109. If using a table, we would identify that it falls between 0.1 and 0.2.

Precise value: 0.0404. If using the table: between 0.02 and 0.05.

Precise value: 0.0266. If using the table: between 0.02 and 0.05.

$X^2=$ $\frac{{(717-743)}^2}{743}+$ $\frac{{(369-338)}^2}{338}+$ $\cdots +$ $\frac{{(10-12)}^2}{12}=$ $4.61$

There are $k=7$ groups, so we use $df=k-1=6$⁠.

False. The chi-square distribution has one parameter called degrees of freedom.

True.

True.

False. As the degrees of freedom increases, the shape of the chi-square distribution becomes more symmetric.

$H_0$⁠: The distribution of the format of the book used by the students follows the professor’s predictions. $H_A$⁠: The distribution of the format of the book used by the students does not follow the professor’s predictions.

$E_{hardcopy}=126\times 0.60=75.6$⁠. $E_{print}=126\times 0.25=31.5$⁠. $E_{online}=126\times 0.15=18.9$⁠.

Independence: The sample is not random. However, if the professor has reason to believe that the proportions are stable from one term to the next and students are not affecting each other’s study habits, independence is probably reasonable. Sample size: All expected counts are at least 5.

${\chi }^2=2.32$⁠, $df=2$⁠, p-value = 0.313.

Since the p-value is large, we fail to reject $H_0$⁠. The data do not provide strong evidence indicating the professor’s predictions were statistically inaccurate.

We would expect $(1-0.2785)\times 73=52.67$⁠. It is okay that this result, like the result from Example 6.38, is a fraction.

The expected count for row one / column one is found by multiplying the row one total (234) and column one total (319), then dividing by the table total (699): $\frac{234\times 319}{699}=106.8$⁠. Similarly for the second column and the first row: $\frac{234\times 380}{699}=127.2$⁠. Row 2: 105.9 and 126.1. Row 3: 106.3 and 126.7.

For each cell, compute $\frac{{(\text{obs}-\text{exp})}^2}{exp}$⁠. For instance, the first row and first column: $\frac{{(109-106.8)}^2}{106.8}=0.05$⁠. Adding the results of each cell gives the chi-square test statistic: $X^2=0.05+\cdots +2.11=8.16$.

If using a computer, we can identify the p-value as 0.017. That is, we reject the null hypothesis because the p-value is less than 0.05, and we conclude that at least one of the treatments is more or less effective than the others at treating Type 2 Diabetes for glycemic control.

(a) Two-way table:

(b-i) $E_{ro{w}_1,co{l}_1}=$ $\frac{(row1total)\times (col1total)}{tabletotal}=$ $35$⁠. This is lower than the observed value.
(b-ii) $E_{ro{w}_2,co{l}_2}=$ $\frac{(row2total)\times (col2total)}{tabletotal}=$ $115$⁠. This is lower than the observed value.

$H_0$⁠: The opinion of college grads and non-grads is not different on the topic of drilling for oil and natural gas off the coast of California. $H_A$⁠: Opinions regarding the drilling for oil and natural gas off the coast of California has an association with earning a college degree.

$$\begin{array}{cccc}\hfill & E_{row1,col1}=151.5\hfill & \hfill & E_{row1,col2}=134.5\hfill \\ \hfill & E_{row2,col1}=162.1\hfill & \hfill & E_{row2,col2}=143.9\hfill \\ \hfill & E_{row3,col1}=124.5\hfill & \hfill & E_{row3,col2}=110.5\hfill \end{array}$$

Independence: The samples are both random, unrelated, and from less than 10% of the population, so independence between observations is reasonable. Sample size: All expected counts are at least 5. ${\chi }^2=11.47$⁠, $df=2$ $\to$ p-value = 0.003. Since the p-value $<\alpha$⁠, we reject $H_0$⁠. There is strong evidence that there is an association between support for off-shore drilling and having a college degree.

No. The samples at the beginning and at the end of the semester are not independent since the survey is conducted on the same students.

$H_0$⁠: The age of Los Angeles residents is independent of shipping carrier preference variable. $H_A$⁠: The age of Los Angeles residents is associated with the shipping carrier preference variable.

The conditions are not satisfied since some expected counts are below 5.

Independence is satisfied (random sample), as is the success-failure condition (40 smokers, 160 non-smokers). The 95% CI: (0.145, 0.255). We are 95% confident that 14.5% to 25.5% of all students at this university smoke.

We want $z^{\star }SE$ to be no larger than 0.02 for a 95% confidence level. We use $z^{\star }=1.96$ and plug in the point estimate $\hat{p}=0.2$ within the SE formula: $1.96\sqrt{0.2(1-0.2)/n}\le 0.02$⁠. The sample size $n$ should be at least 1,537.

Proportion of graduates from this university who found a job within one year of graduating. $\hat{p}=348/400=0.87$⁠.

This is a random sample, so the observations are independent. Success-failure condition is satisfied: 348 successes, 52 failures, both well above 10.

(0.8371, 0.9029). We are 95% confident that approximately 84% to 90% of graduates from this university found a job within one year of completing their undergraduate degree.

95% of such random samples would produce a 95% confidence interval that includes the true proportion of students at this university who found a job within one year of graduating from college.

(0.8267, 0.9133). Similar interpretation as before.

99% CI is wider, as we are more confident that the true proportion is within the interval and so need to cover a wider range.

Use a chi-squared goodness of fit test. $H_0$⁠: Each option is equally likely. $H_A$⁠: Some options are preferred over others. Total sample size: 99. Expected counts: (1/3) * 99 = 33 for each option. These are all above 5, so conditions are satisfied. $df=3-1=2$ and ${\chi }^2=$ $\frac{{(43-33)}^2}{33}+$ $\frac{{(21-33)}^2}{33}+$ $\frac{{(35-33)}^2}{33}=$ $7.52\to$ p-value $=0.023$⁠. Since the p-value is less than 5%, we reject $H_0$⁠. The data provide convincing evidence that some options are preferred over others.

$H_0:p=0.38$⁠. $H_A:p\ne 0.38$⁠. Independence (random sample) and the success-failure condition are satisfied. $Z=-20.5$ $\to$ p-value $\approx 0$⁠. Since the p-value is very small, we reject $H_0$⁠. The data provide strong evidence that the proportion of Americans who only use their cell phones to access the internet is different than the Chinese proportion of 38%, and the data indicate that the proportion is lower in the US.

If in fact 38% of Americans used their cell phones as a primary access point to the internet, the probability of obtaining a random sample of 2,254 Americans where 17% or less or 59% or more use their only their cell phones to access the internet would be approximately 0.

(0.1545, 0.1855). We are 95% confident that approximately 15.5% to 18.6% of all Americans primarily use their cell phones to browse the internet.