7.58

Physical demands of women’s rugby seven matches. Rugby sevens is rapidly growing in popularity and will be included in the 2016 Olympics. Matches are played on a full rugby field and consist of two seven-minute halves. Each team also consists of seven players. To better understand the demands of women’s rugby sevens, a group of researchers compared the physical qualities of elite players from the Canadian National team with a university squad.
Carry out the significance tests using α=0.05α=0.05. Report the test statistic with the degrees of freedom and the P-value. Write a short summary of your conclusion.

In this case, the null hypothesis states that there is no difference between the elite rugby players and the university players, and the alternative hypothesis states that there is a difference between two groups. Using the formula for the two-tailed test, I summed up all the mean scores and the SD from these two groups and calcuate the test statistics using the calculator. The result shows that t = 1.95, and p-value equals 0.0611. Given the alpha 0.05, we fail to reject the null hypothesis. Therefore, we cannot conclude that there is a significant difference between the physical qualities of the university players and the elite players.

7.68

Diet and mood. Researchers were interested in comparing the long-term psychological effects of being on a high-carbohydrate, low-fat (LF) diet versus a high-fat, low-carbohydrate (LC) diet. A total of 106 overweight and obese participants were randomly assigned to one of these two energy-restricted diets. At 52 weeks, 32 LC dieters and 33 LF dieters remained. Mood was assessed using a total mood disturbance score (TMDS).
(a) Is there a difference in the TMDS at Week 52? Test the null hypothesis that the dieters’ average mood in the two groups is the same. Use a significance level of 0.05.

Given the mean and the SD in the summary table. We calculate the test statistics using the calculator and the result shows that p = 0.0001. Given this number, we can reject the null hypothesis in favor of the alternative hypothesis that there is a difference in the TMDS between these two groups.

(b) Critics of this study focus on the specific LC diet (that it, the science) and the dropout rate. Explain why the dropout rate is important to consider when drawing conclusions from this study.

The dropout rate assesses the rate in which a person responses to the certain diet. Throughout the 52 weeks, people who have high dropout rates will response faster to the diet, with others with low dropout rate cannot change their weights quickly. Therefore, considering this dropout rate is important in drawing conclusion from the study.

7.80

Size of trees in the eastern and western halves. Refer to the previous exercise. The Wade Tract can also be divided into eastern and western halves. Here are the DBHs of 30 randomly selected longleaf pine trees from each half:
(c) What are appropriate null and alternative hypotheses for comparing the two samples of tree DBHs? Give reasons for your choices.

Null hypothesis: there is no difference between the eastern trees and the western trees. \[Ho: \mu_1 - \mu_2 = 0\]

Alternative hypothesis: there are differences between the eastern trees and the western trees. \[Ha: \mu_1 - \mu_2 ≠ 0 \]

# divide the dataset into two groups:
treeW <- tree %>% 
  filter(ew == "w") %>% 
  group_by(ew) 

treeE <- tree %>% 
  filter(ew == "e") %>% 
  group_by(ew)

From these results, we use the test statistics formula for two tailed test, the result is as followed:

t.test(treeE$dbh, treeW$dbh, mu = 0, conf.level = 0.95, alternative = "two.sided", paired = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  treeE$dbh and treeW$dbh
## t = -2.1123, df = 57.871, p-value = 0.03899
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -16.6852180  -0.4481154
## sample estimates:
## mean of x mean of y 
##  21.71667  30.28333
(d) Perform the significance test. Report the test statistic, the degrees of freedom, and the P-value. Summarize your conclusion.

The table shows t = -2.1123, with the degree of freedom df = 57.8. The p-value equals 0.03. With this given p-value, we can reject the null hypothesis in favor of the alternative hypothesis saying that there is a difference between the eastern halves and the western halves.

(e) Find a 95% confidence interval for the difference in mean DBHs. Explain how this interval provides additional information about this problem.
# the mean difference is:
diff <- treeE$dbh - treeW$dbh

t.test(diff)
## 
##  One Sample t-test
## 
## data:  diff
## t = -2.01, df = 29, p-value = 0.05382
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  -17.2835364   0.1502031
## sample estimates:
## mean of x 
## -8.566667

The confidence interval is between -17.2 and 0.15. This shows that 95% of the true population mean differences between these two trees are in this intervals.

7.82

An improper significance test. A friend has performed a significance test of the null hypothesis that two means are equal. His report states that the null hypothesis is rejected in favor of the alternative that the first mean is larger than the second. In a presentation on his work, he notes that the first sample mean was larger than the second mean and this is why he chose this particular one-sided alternative.

(a) Explain what is wrong with your friend’s procedure and why.

This should be a two-tailed test. This is because if the null hypothesis assumes that two means are equal, then the alternative hypothesis should state two means are not equal. In here he chooses the one-sided alternative, so it is not an appropiate test.

(b) Suppose that he reported t=1.93 with a P-value of 0.06. What is the correct P-value that he should report?

Since he performed the one-sided test, the p-value is 0.06. If he performed the two tailed test, then the p value should be 0.06*2 = 0.12.

7.123

Which design? The following situations all require inference about a mean or means. Identify each as (1) a single sample, (2) matched pairs, or (3) two independent samples. Explain your answers.

(a) Your customers are college students. You are interested in comparing the interest in a new product that you are developing between those students who live in the dorms and those who live elsewhere.

This will have two independent samples. The first sample is those who live in the dorms and the other live elsewhere.

(b) Your customers are college students. You are interested in finding out which of two new product labels is more appealing.

This is a matched pairs design. We compare the response of only college students testing the two new product lables.

(c) Your customers are college students. You are interested in assessing their interest in a new product.

This is a single sample design. The sample in here is college students testing a new product.

7.124

Which design? The following situations all require inference about a mean or means. Identify each as (1) a single sample, (2) matched pairs, or (3) two independent samples. Explain your answers.

(a) You want to estimate the average age of your store’s customers.

This is a single sample design. The only sample we need here is the store’s customers.

(b) You do an SRS survey of your customers every year. One of the questions on the survey asks about customer satisfaction on a seven-point scale with the response 1 indicating “very dissatisfied’’ and 7 indicating “very satisfied.’’ You want to see if the mean customer satisfaction has improved from last year.

This is the two independent samples design. In this case we compare the customer satistisfaction of two groups of customers. The first group is the customers from last year and the other is the group from this year.

(c) You ask an SRS of customers their opinions on each of two new floor plans for your store.

We have a matched-pair design in this case because one group of people gave opinions on two new floor plans.