Chapter 7: Inference for numerical dataContent check — Statistics (OpenIntro), Chapter 7: Inference for numerical data

We want to find the shaded area above -1.79 (we leave the picture to you). The lower tail area has an area of 0.0447, so the upper area would have an area of $1-0.0447=0.9553$⁠.

The sample size is under 30, so we check for obvious outliers: since all observations are within 2 standard deviations of the mean, there are no such clear outliers.

$\bar{x}\pm t_{14}^{\star }\times SE\to$ $0.287\pm 1.76\times 0.0178\to$ $(0.256,0.318)$⁠. We are 90% confident that the average mercury content of croaker white fish (Pacific) is between 0.256 and 0.318 ppm.

No, a confidence interval only provides a range of plausible values for a population parameter, in this case the population mean. It does not describe what we might observe for individual observations.

$H_0$⁠: The average 10-mile run time was the same for 2006 and 2017. $\mu =93.29$ minutes. $H_A$⁠: The average 10-mile run time for 2017 was different than that of 2006. $\mu \ne 93.29$ minutes.

With a sample of 100, we should only be concerned if there is are particularly extreme outliers. The histogram of the data doesn’t show any outliers of concern (and arguably, no outliers at all).

$df=6-1=5$⁠, $t_5^{\star }=2.02$ (column with two tails of 0.10, row with $df=5$⁠).

$df=21-1=20$⁠, $t_{20}^{\star }=2.53$ (column with two tails of 0.02, row with $df=20$⁠).

$df=28$⁠, $t_{28}^{\star }=2.05$⁠.

$df=11$⁠, $t_{11}^{\star }=3.11$⁠.

0.085, do not reject $H_0$⁠.

0.003, reject $H_0$⁠.

0.438, do not reject $H_0$⁠.

0.042, reject $H_0$⁠.

The mean is the midpoint: $\bar{x}=20$⁠. Identify the margin of error: $ME=1.015$⁠, then use $t_{35}^{\star }=2.03$ and $SE=s/\sqrt{n}$ in the formula for margin of error to identify $s=3$⁠.

$H_0$⁠: $\mu =8$ (New Yorkers sleep 8 hrs per night on average.) $H_A$⁠: $\mu \ne 8$ (New Yorkers sleep less or more than 8 hrs per night on average.)

Independence: The sample is random. The min/max suggest there are no concerning outliers. $T=-1.75$⁠. $df=25-1=24$⁠.

p-value $=0.093$⁠. If in fact the true population mean of the amount New Yorkers sleep per night was 8 hours, the probability of getting a random sample of 25 New Yorkers where the average amount of sleep is 7.73 hours per night or less (or 8.27 hours or more) is 0.093.

Since p-value $>$ 0.05, do not reject $H_0$⁠. The data do not provide strong evidence that New Yorkers sleep more or less than 8 hours per night on average.

No, since the p-value is smaller than $1-0.90=0.10$⁠.

$T$ is either -2.09 or 2.09. Then $\bar{x}$ is one of the following:

$$\begin{array}{cc}\hfill -2.09& =\frac{\bar{x}-60}{\frac{8}{\sqrt{20}}}\to \bar{x}=56.26\hfill \\ \hfill 2.09& =\frac{\bar{x}-60}{\frac{8}{\sqrt{20}}}\to \bar{x}=63.74\hfill \end{array}$$

We will conduct a 1-sample $t$⁠-test. $H_0$⁠: $\mu =5$⁠. $H_A$⁠: $\mu \ne 5$⁠. We’ll use $\alpha =0.05$⁠. This is a random sample, so the observations are independent. To proceed, we assume the distribution of years of piano lessons is approximately normal. $SE=2.2/\sqrt{20}=0.4919$⁠. The test statistic is $T=(4.6-5)/SE=-0.81$⁠. $df=20-1=19$⁠. The one-tail area is about 0.21, so the p-value is about 0.42, which is bigger than $\alpha =0.05$ and we do not reject $H_0$⁠. That is, we do not have sufficiently strong evidence to reject the notion that the average is 5 years.

Using $SE=0.4919$ and $t_{df=19}^{\star }=2.093$⁠, the confidence interval is (3.57, 5.63). We are 95% confident that the average number of years a child takes piano lessons in this city is 3.57 to 5.63 years.

They agree, since we did not reject the null hypothesis and the null value of 5 was in the $t$⁠-interval.

If the sample is large, then the margin of error will be about $1.96\times 100/\sqrt{n}$⁠. We want this value to be less than 10, which leads to $n\ge 384.16$⁠, meaning we need a sample size of at least 385 (round up for sample size calculations!).

Conditions have already verified and the standard error computed in Example 7.17. To find the interval, identify $t_{67}^{\star }$ using statistical software or the $t$⁠-table ($t_{67}^{\star }=2.00$⁠), and plug it, the point estimate, and the standard error into the confidence interval formula:

$$\begin{array}{c}\hfill \text{point estimate}\pm z^{\star }\times SE\to 3.58\pm 2.00\times 1.63\to (0.32,6.84)\end{array}$$

We are 95% confident that Amazon is, on average, between $0.32 and $6.84 less expensive than the UCLA Bookstore for UCLA course books.

The average price difference is only mildly useful for this question. Examine the distribution shown in Figure 7.9. There are certainly a handful of cases where Amazon prices are far below the UCLA Bookstore’s, which suggests it is worth checking Amazon (and probably other online sites) before purchasing. However, in many cases the Amazon price is above what the UCLA Bookstore charges, and most of the time the price isn’t that different. Ultimately, if getting a book immediately from the bookstore is notably more convenient, e.g. to get started on reading or homework, it’s likely a good idea to go with the UCLA Bookstore unless the price difference on a specific book happens to be quite large.

For reference, this is a very different result from what we (the authors) had seen in a similar data set from 2010. At that time, Amazon prices were almost uniformly lower than those of the UCLA Bookstore’s and by a large margin, making the case to use Amazon over the UCLA Bookstore quite compelling at that time. Now we frequently check multiple websites to find the best price.

Paired, data are recorded in the same cities at two different time points. The air quality in a city at one point is not independent of the air quality in the same city at another time point.

Since it’s the same students at the beginning and the end of the semester, there is a pairing between the datasets, for a given student their beginning and end of semester grades are dependent.

Since the subjects were sampled randomly, each observation in the men’s group does not have a special correspondence with exactly one observation in the other (women’s) group.

Since it’s the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester artery thickness are dependent.

Since it’s the same subjects at the beginning and the end of the study, there is a pairing between the datasets, for a subject student their beginning and end of semester weights are dependent.

For each observation in one data set, there is exactly one specially corresponding observation in the other data set for the same geographic location. The data are paired.

$H_0:{\mu }_{\text{diff}}=0$ (There is no difference in average number of days exceeding 90°F in 1948 and 2018 for NOAA stations.) $H_A:{\mu }_{\text{diff}}\ne 0$ (There is a difference.)

Locations were randomly sampled, so independence is reasonable. The sample size is at least 30, so we’re just looking for particularly extreme outliers: none are present (the observation off left in the histogram would be considered a clear outlier, but not a particularly extreme one). Therefore, the conditions are satisfied.

$SE=17.2/\sqrt{197}=1.23$⁠. $T=\frac{2.9-0}{1.23}=2.36$ with degrees of freedom $df=197-1=196$⁠. This leads to a one-tail area of 0.0096 and a p-value of about 0.019.

Since the p-value is less than 0.05, we reject $H_0$⁠. The data provide strong evidence that NOAA stations observed more 90°F days in 2018 than in 1948.

Type 1 Error, since we may have incorrectly rejected $H_0$⁠. This error would mean that NOAA stations did not actually observe a decrease, but the sample we took just so happened to make it appear that this was the case.

No, since we rejected $H_0$⁠, which had a null value of 0.

$SE=1.23$ and $t^{\star }=1.65$⁠. $2.9\pm 1.65\times 1.23\to$ $(0.87,4.93)$⁠.

We are 90% confident that there was an increase of 0.87 to 4.93 in the average number of days that hit 90°F in 2018 relative to 1948 for NOAA stations.

Yes, since the interval lies entirely above 0.

The difference in sample means is an appropriate point estimate: ${\bar{x}}_n-{\bar{x}}_s=0.40$⁠.

The standard error of the estimate can be calculated using the standard error formula:

$$\begin{array}{c}\hfill SE=\sqrt{\frac{{\sigma }_n^2}{n_n}+\frac{{\sigma }_s^2}{n_s}}\approx \sqrt{\frac{s_n^2}{n_n}+\frac{s_s^2}{n_s}}=\sqrt{\frac{{1.60}^2}{100}+\frac{{1.43}^2}{50}}=0.26\end{array}$$

It is possible that there is a difference but we did not detect it. If there is a difference, we made a Type 2 Error.

We could have collected more data. If the sample sizes are larger, we tend to have a better shot at finding a difference if one exists. In fact, this is exactly what we would find if we examined a larger data set!

$H_0$⁠: the exams are equally difficult, on average. ${\mu }_A-{\mu }_B=0$⁠. $H_A$⁠: one exam was more difficult than the other, on average. ${\mu }_A-{\mu }_B\ne 0$⁠.

Since the exams were shuffled, the “treatment” in this case was randomly assigned, so independence within and between groups is satisfied.

The summary statistics suggest the data are roughly symmetric about the mean, and the min/max values don’t suggest any outliers of concern.

These data are paired. For example, the Friday the 13th in say, September 1991, would probably be more similar to the Friday the 6th in September 1991 than to Friday the 6th in another month or year.

Let ${\mu }_{\text{𝑑𝑖𝑓𝑓}}=$ ${\mu }_{sixth}-$ ${\mu }_{thirteenth}$⁠. $H_0:{\mu }_{\text{𝑑𝑖𝑓𝑓}}=0$⁠. $H_A:{\mu }_{\text{𝑑𝑖𝑓𝑓}}\ne 0$⁠.

Independence: The months selected are not random. However, if we think these dates are roughly equivalent to a simple random sample of all such Friday 6th/13th date pairs, then independence is reasonable. To proceed, we must make this strong assumption, though we should note this assumption in any reported results. Normality: With fewer than 10 observations, we would need to see clear outliers to be concerned. There is a borderline outlier on the right of the histogram of the differences, so we would want to report this in formal analysis results.

$T=4.93$ for $df=10-1=9$ $\to$ p-value = 0.001.

Since p-value $<$ 0.05, reject $H_0$⁠. The data provide strong evidence that the average number of cars at the intersection is higher on Friday the 6⁠${}^{\text{th}}$ than on Friday the 13⁠${}^{\text{th}}$⁠. (We should exercise caution about generalizing the interpretation to all intersections or roads.)

If the average number of cars passing the intersection actually was the same on Friday the 6⁠${}^{\text{th}}$ and ${13}^{th}$⁠, then the probability that we would observe a test statistic so far from zero is less than 0.01.

We might have made a Type 1 Error, i.e. incorrectly rejected the null hypothesis.

$H_0:{\mu }_{diff}=0$⁠. $H_A:{\mu }_{diff}\ne 0$⁠. $T=-2.71$⁠. $df=5$⁠. p-value $=0.042$⁠. Since p-value $<$ 0.05, reject $H_0$⁠. The data provide strong evidence that the average number of traffic accident related emergency room admissions are different between Friday the 6⁠${}^{\text{th}}$ and Friday the 13⁠${}^{\text{th}}$⁠. Furthermore, the data indicate that the direction of that difference is that accidents are lower on Friday the $6^{th}$ relative to Friday the 13⁠${}^{\text{th}}$⁠.

(-6.49, -0.17).

This is an observational study, not an experiment, so we cannot so easily infer a causal intervention implied by this statement. It is true that there is a difference. However, for example, this does not mean that a responsible adult going out on Friday the ${13}^{th}$ has a higher chance of harm than on any other night.

Chicken fed linseed weighed an average of 218.75 grams while those fed horsebean weighed an average of 160.20 grams. Both distributions are relatively symmetric with no apparent outliers. There is more variability in the weights of chicken fed linseed.

$H_0:{\mu }_{ls}={\mu }_{hb}$⁠. $H_A:{\mu }_{ls}\ne {\mu }_{hb}$⁠.
We leave the conditions to you to consider.
$T=3.02$⁠, $df=min(11,9)=9$ $\to$ p-value $=0.014$⁠. Since p-value $<$ 0.05, reject $H_0$⁠. The data provide strong evidence that there is a significant difference between the average weights of chickens that were fed linseed and horsebean.

Type 1 Error, since we rejected $H_0$⁠.

Yes, since p-value $>$ 0.01, we would not have rejected $H_0$⁠.

$H_0:{\mu }_C={\mu }_S$⁠. $H_A:{\mu }_C\ne {\mu }_S$⁠. $T=3.27$⁠, $df=11$ $\to$ p-value $=0.007$⁠. Since p-value $<0.05$⁠, reject $H_0$⁠. The data provide strong evidence that the average weight of chickens that were fed casein is different than the average weight of chickens that were fed soybean (with weights from casein being higher). Since this is a randomized experiment, the observed difference can be attributed to the diet.

Let ${\mu }_{diff}={\mu }_{pre}-{\mu }_{post}$⁠. $H_0:{\mu }_{diff}=0$⁠: Treatment has no effect. $H_A:{\mu }_{diff}\ne 0$⁠: Treatment has an effect on P.D.T. scores, either positive or negative. Conditions: The subjects are randomly assigned to treatments, so independence within and between groups is satisfied. All three sample sizes are smaller than 30, so we look for clear outliers. There is a borderline outlier in the first treatment group. Since it is borderline, we will proceed, but we should report this caveat with any results. For all three groups: $df=13$⁠. $T_1=1.89\to$ p-value = 0.081, $T_2=1.35\to$ p-value = 0.200), $T_3=-1.40\to$ (p-value = 0.185). We do not reject the null hypothesis for any of these groups. As earlier noted, there is some uncertainty about if the method applied is reasonable for the first group.

The standard error is given as $SE=\sqrt{\frac{{12}^2}{500}+\frac{{12}^2}{500}}=0.76$⁠.

&

The null distribution, rejection boundaries, and alternative distribution are shown below:

The rejection regions are the areas on the outside of the two dotted lines and are at $\pm 0.76\times 1.96=\pm 1.49$⁠.

The area of the alternative distribution where ${\mu }_{trmt}-{\mu }_{ctrl}=-3$ has been shaded. We compute the Z-score and find the tail area: $Z=$ $\frac{-1.49-(-3)}{0.76}=$ $1.99\to$ $0.977$⁠. With 500 patients per group, we would be about 97.7% sure (or more) that we’d detect any effects that are at least 3 mmHg in size.

First, find the Z-score such that 90% of the distribution is below it: $Z=1.28$⁠. Next, find the cutoffs for the rejection regions: $\pm 2.58$⁠. Then the difference in centers should be about $1.28\times SE+2.58\times SE=3.86\times SE$⁠.

Answers will vary, but here are a few important considerations:

• Whether there is any risk to patients in the study.

• The cost of enrolling more patients.

• The potential downside of not detecting an effect of interest.

Difference we care about: 40. Single tail of 90%: $1.28\times SE$⁠. Rejection region bounds: $\pm 1.96\times SE$ (if 5% significance level). Setting $3.24\times SE=40$⁠, subbing in $SE=\sqrt{\frac{{94}^2}{n}+\frac{{94}^2}{n}}$⁠, and solving for the sample size $n$ gives 116 plots of land for each fertilizer.

$H_0$⁠: The average on-base percentage is equal across the three positions. $H_A$⁠: The average on-base percentage varies across some (or all) groups.

There are $k=3$ groups, so $d{f}_G=k-1=2$⁠. There are $n=n_1+n_2+n_3=429$ total observations, so $d{f}_E=n-k=426$⁠. Then the $F$ statistic is computed as the ratio of $MSG$ and $MSE$⁠: $F=$ $\frac{MSG}{MSE}=$ $\frac{0.00803}{0.00158}=$ $5.082\approx$ $5.077$⁠. ($F=5.077$ was computed by using values for $MSG$ and $MSE$ that were not rounded.)

The p-value of the test is 0.0330, less than the default significance level of 0.05. Therefore, we reject the null hypothesis and conclude that the difference in the average midterm scores are not due to chance.

Alternative.

$H_0$⁠: ${\mu }_1={\mu }_2=\cdots ={\mu }_6$⁠. $H_A$⁠: The average weight varies across some (or all) groups. Independence: Chicks are randomly assigned to feed types (presumably kept separate from one another), therefore independence of observations is reasonable. Approx. normal: the distributions of weights within each feed type appear to be fairly symmetric. Constant variance: Based on the side-by-side box plots, the constant variance assumption appears to be reasonable. There are differences in the actual computed standard deviations, but these might be due to chance as these are quite small samples. $F_{\mathrm{5,65}}=15.36$ and the p-value is approximately 0. With such a small p-value, we reject $H_0$⁠. The data provide convincing evidence that the average weight of chicks varies across some (or all) feed supplement groups.

$H_0$⁠: The population mean of MET for each group is equal to the others. $H_A$⁠: At least one pair of means is different.

Independence: We don’t have any information on how the data were collected, so we cannot assess independence. To proceed, we must assume the subjects in each group are independent. In practice, we would inquire for more details. Normality: The data are bound below by zero and the standard deviations are larger than the means, indicating very strong skew. However, since the sample sizes are extremely large, even extreme skew is acceptable. Constant variance: This condition is sufficiently met, as the standard deviations are reasonably consistent across groups.

See below, with the last column omitted:

Since p-value is very small, reject $H_0$⁠. The data provide convincing evidence that the average MET differs between at least one pair of groups.

$H_0$⁠: Average GPA is the same for all majors. $H_A$⁠: At least one pair of means are different.

Since p-value $>$ 0.05, fail to reject $H_0$⁠. The data do not provide convincing evidence of a difference between the average GPAs across three groups of majors.

The total degrees of freedom is $195+2=197$⁠, so the sample size is $197+1=198$⁠.

False. As the number of groups increases, so does the number of comparisons and hence the modified significance level decreases.

True.

True.

False. We need observations to be independent regardless of sample size.

$H_0$⁠: Average score difference is the same for all treatments. $H_A$⁠: At least one pair of means are different.

We should check conditions. If we look back to the earlier exercise, we will see that the patients were randomized, so independence is satisfied. There are some minor concerns about skew, especially with the third group, though this may be acceptable. The standard deviations across the groups are reasonably similar. Since the p-value is less than 0.05, reject $H_0$⁠. The data provide convincing evidence of a difference between the average reduction in score among treatments.

We determined that at least two means are different in part (b), so we now conduct $K=3\times 2/2=3$ pairwise $t$⁠-tests that each use $\alpha =0.05/3=0.0167$ for a significance level. Use the following hypotheses for each pairwise test. $H_0$⁠: The two means are equal. $H_A$⁠: The two means are different. The sample sizes are equal and we use the pooled SD, so we can compute $SE=3.7$ with the pooled $df=39$⁠. Looking at the largest difference, Trmt 1 vs Trmt 3: $Z=\frac{6.21-(-3.21)}{3.7}=2.52$ on $df=39$ yields a p-value of 0.015. Because this is smaller than $0.05/3=1.67$⁠, we have strong evidence to that this particular pair of groups are different. When doing similar calculations for Trmt 1 vs 2 or 2 vs 3, we do not find any statistically significant difference. (Note that we get a different result if not using the pooled result.)

$H_0:{\mu }_T={\mu }_C$⁠. $H_A:{\mu }_T\ne {\mu }_C$⁠. $T=2.24$⁠, $df=21$ $\to$ p-value $=0.036$⁠. Since p-value $<$ 0.05, reject $H_0$⁠. The data provide strong evidence that the average food consumption by the patients in the treatment and control groups are different. Furthermore, the data indicate patients in the distracted eating (treatment) group consume more food than patients in the control group.

False. While it is true that paired analysis requires equal sample sizes, only having the equal sample sizes isn’t, on its own, sufficient for doing a paired test. Paired tests require that there be a special correspondence between each pair of observations in the two groups.

We are building a distribution of sample statistics, in this case the sample mean. Such a distribution is called a sampling distribution.

Because we are dealing with the distribution of sample means, we need to check to see if the Central Limit Theorem applies. Our sample size is greater than 30, and we are told that random sampling is employed. With these conditions met, we expect that the distribution of the sample mean will be nearly normal and therefore symmetric.

Because we are dealing with a sampling distribution, we measure its variability with the standard error. $SE=18.2/\sqrt{45}=2.713$⁠.

The sample means will be more variable with the smaller sample size.

We should set 1.0% equal to 2.8 standard errors: $2.8\times S{E}_{desired}=1.0\%$ (see Example 7.37 on page  for details). This means the standard error should be about $SE=0.36\%$ to achieve the desired statistical power.

The margin of error was $0.5\times (2.6\%-(-0.2\%))=1.4\%$⁠, so the standard error in the experiment must have been $1.96\times S{E}_{original}=1.4\%$ $\to$ $S{E}_{original}=0.71\%$⁠.

The standard error decreases with the square root of the sample size, so we should increase the sample size by a factor of ${1.97}^2=3.88$⁠.

The team should run an experiment 3.88 times larger, so they should have a random sample of 3.88% of their users in each of the experiment arms in the new experiment.

Independence: it is a random sample, so we can assume that the students in this sample are independent of each other with respect to number of exclusive relationships they have been in. Notice that there are no students who have had no exclusive relationships in the sample, which suggests some student responses are likely missing (perhaps only positive values were reported). The sample size is at least 30, and there are no particularly extreme outliers, so the normality condition is reasonable. 90% CI: (2.97, 3.43). We are 90% confident that undergraduate students have been in 2.97 to 3.43 exclusive relationships, on average.

The hypotheses should be about the population mean ($\mu$⁠), not the sample mean. The null hypothesis should have an equal sign and the alternative hypothesis should be about the null hypothesized value, not the observed sample mean. Correction:

$$\begin{array}{cc}\hfill H_0& :\mu =10hours\hfill \\ \hfill H_A& :\mu \ne 10hours\hfill \end{array}$$

A two-sided test allows us to consider the possibility that the data show us something that we would find surprising.