Week 7 Discussion Boarda. Answer the following question, making sure to explain your thinking in detail.

AL baseball teams play with a designated hitter, meaning pitchers don’t bat. The league believes

that replacing the pitcher, typically a weak hitter, with another player produces more runs and

generates more interest among fans. Below are the average numbers of runs scored in AL and

NL stadiums for the 2006 season.

1. With a 99% confidence interval, estimate the mean number of runs scored in AL games.

2. Coors Field, in Denver, stands a mile above sea level, and some believe the thinner air

makes it harder for pitchers to throw curve balls and easier for batters to hit the ball far.

Do you think the 10.5 runs scored per game at Coors Field is unusual? Explain.

b. Find one aspect of this week’s material that is relevant to college, career, or everyday life.

Provide some detail on how it could be important.

Week 7 Problem Set:

1.

Some archaeologists believe that ancient Egyptians interbred with several

different immigrant populations over thousands of years. To see if there is any

indication of changes in body structure as a result, they measured 30 skulls of

male Egyptians dated from 4000 BCE and 30 others dated from 200 BCE.

a. Are these data appropriate for inference? Explain.

b. Create a 90% confidence interval for the difference in mean skull breadth

between these two eras.

c. Do these data provide evidence that the mean breadth of males’ skulls

changed over this period? Explain.

2.

3.

Researchers investigated how the size of a bowl impacts how much ice cream

people scoop. People were randomly given a 17 oz or a 34 oz bowl. They were

then invited to scoop as much ice cream as they liked. Did the bowl size change

the selected portion size? Test an appropriate hypothesis and state your

conclusions.

Every year students at a certain high school take a physical fitness test during

gym. One component asks them to do as many push-ups as they can. Results for

one class are shown below, separately for boys and girls. Assuming that students

are assigned to gym classes at random, create a 95% confidence interval for how

many more push-ups boys can do than girls, on average, at that high school.

4.

A company institutes an exercise break for its workers to see if it will improve

job satisfaction (as measured by a questionnaire). Scores for 10 randomly

selected workers before and after implementation of the program are shown in

the table below.

a. Identify the procedure you would use to assess the effectiveness of the

exercise program.

b. Test an appropriate hypothesis and state your conclusion.

Just like we did with proportions, we are now going to discuss how to compare means. In

Chapter 24, we learn about two-sample t-tests. When comparing two means, we are concerned

with the difference between the two observed means, 𝑦̅1 − 𝑦̅2 , as well as the standard error,

represented by

The standard error is going to help us

determine our two-sample t-interval, which should look familiar:

(𝑦̅1 − 𝑦̅2 ) ± 𝑀𝐸, where ME = t* x SE(𝑦̅1 − 𝑦̅2 ). This formula should look just like the formula

for a confidence interval for proportions, except here we are using a Student’s t model, rather

than a normal model. Here is the sampling distribution information, for convenience.

Please note assumptions and conditions on pp. 563-564. When trying to find a two-sample tinterval, we can use the following to guide us.

Now let’s try an example.

a. The boxplots suggest that male and female pulse rates are not all that different, but that

female pulse rates have a slightly larger spread.

b. Using the assumptions and conditions references on pp. 563-564, we can say that the

groups are independent, we can assume a random, representative sample, and one can see

that the boxplots are reasonably symmetric.

c. Let’s create a 90% confidence interval. The observed difference in means is 72.75 –

72.625 = 0.125. Degrees of freedom is harder to come by, and I know the textbook

suggests a more complicated formula, but since this is an intro course, let’s use the easier

formula, or n – 1 from the lower n. Therefore, our df = 24 – 1 = 23. In order to find t*23,

we use the function T.INV.2T() on Excel, which calls for an alpha and df. Therefore,

=T.INV.2T(.1,23) = 1.7138. Finally, we need the standard error, which we get with the

complicated formula above. Calculating sqrt(5.372252/28 + 7.699872/24), we get a SE of

1.8711. Now, putting it all together, we get 0.125 plus/minus 1.7138 * 1.8711, which

gives us an interval of (-3.08, 3.33).

d. Yes. The confidence interval clearly includes a possible mean difference of 0, which

would indicate that there is no evidence of a difference in pulse rate for men and women.

Hopefully this is starting to feel familiar. Once again, we are taking a statistic plus or minus a

margin of error. Furthermore, this margin of error is based on a critical value times the standard

error. Similar to the work we did with proportions, then, let’s take a look at a hypothesis test for

the difference between two means.

Let’s do another example.

Let’s start with our hypotheses. Our null hypothesis would state that there is no difference

between seniors’ science scores in 2000 and 1996. Our alternative hypothesis would state that

the scores in 2000 were less than in 1996.

a. The observed difference is 3 points. The standard error is 1.22. The degrees of freedom

would be (again, cheating a bit) 7537 – 1 = 7536. Our t statistic, then, can be calculated

using the formula above, which gives us 3/1.22 = 2.46. Now let’s run a t-test with

technology. The command on Excel is =TDIST(), and we have to input our t-statistic,

degrees of freedom, and number of tails. For this problem, we would enter

=TDIST(2.46,7536,1), and we would get .007, which leads us to reject the null. We can

say, then, that there is strong evidence that the mean score on the NAEP science test

decreased from 1996 to 2000.

b. Both sample sizes are large, resulting in very small standard errors. The difference in

sample size should make much of a difference.

Chapter 25 is our final chapter with Students’ t, and here, we consider data that are not

independent. The strategy here is to determine when data are paired, and then we simply find the

differences for each pair, and treat these differences as if they were a one-sample t-test. We will

call this a paired t-test. Assumptions and conditions mirror what we looked at previously (see. P.

589 – 590), but include the requirement that the data are, in fact, paired. See the steps for

conducting a paired t-test below.

Let’s do an example.

The null hypothesis states that the mean difference in calories between servings of strawberry

and vanilla yogurt is zero. The alternative hypothesis states that the mean difference in calories

between servings of strawberry and vanilla yogurt is not zero. The paired data assumption checks

out, we assume randomization, we will assume the 10% condition (although, to be honest, I’m

not sure how many brands of yogurt there are), and we can proceed with the normal population

assumption, if we remove the “Great Value” outlier.

We now proceed with a Student’s t-model with 11 – 1 = 10 degrees of freedom. Now, we find

the differences between each of the pairs, and average those differences (remember –we are

excluding “Great Value”). That average difference is 4.55. The sample standard deviation of

those eleven differences is 18.09 (using =STDEV() on Excel), and we can find the SE by

calculating 18.09/sqrt(n) = 18.09/sqrt(11) = 5.45. Finally, using the formula above, we get t =

4.55/5.45 = 0.833. Finally, we can obtain our P-value in Excel, using =TDIST(0.833,11,2), and

we get 0.42 (remember, =TDIST() takes our t-statistic, df, and number of tails as its inputs).

Since the P-value = 0.42, we fail to reject the null hypothesis. There is no evidence of a mean

difference in calorie content between strawberry yogurt and vanilla yogurt.

Similarly, we can also create a confidence interval for matched pairs, using the following steps:

Let’s do an example here.

Checking assumptions and conditions, the data are paired, the cities may not be representative of

all European cities (so we should be careful about generalizing), the 10% condition checks out,

and we can assume a normal population based on a roughly unimodal and symmetric histogram.

Let’s move forward with a Student’s t-model with 12 – 1 = 11 degrees of freedom. Once again,

we start with the twelve differences, and find the average of those differences (36.833) and the

standard deviation of those differences (8.664). Finally, we need t*11, which we can get with

Excel. I will type in =TINV(.1,11), and I get 1.796 (note, =TINV() takes the alpha level and the

df). Now, I can use my confidence interval formula, 36.833 plus/minus 1.796*8.664/sqrt(12),

which gives me (32.3, 41.3).

In conclusion, we are 90% confident that the average high temperature in European cities in July

is an average of between 32.3 degrees to 41.4 degrees higher than in January.

The text does a nice job demonstrating how the confidence interval can give us some insight into

the effect size. Confidence intervals are really powerful tools, and if you really understand them,

they can be your go-to for quite a bit of the inferential statistics we have been studying.

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