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Twenty-nine college students, identified as having a positive attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were asked to rate how trustworthy the face of Mitt Romney appeared, as represented in their mental image of Mitt Romney's face. Ratings were on a scale of 0 to 7 , with 0 being "not at all trustworthy" and 7 being "extremely trustworthy." Here are the 29 ratings: \({ }^{22}\) \(\begin{array}{llllllllll}2.6 & 3.2 & 3.7 & 3.3 & 3.4 & 3.6 & 3.7 & 3.8 & 3.9 & 4.1\end{array}\) \(\begin{array}{lllllllllll}4.2 & 4.9 & 5.7 & 4.2 & 3.9 & 3.2 & 4.5 & 5.0 & 5.0 & 4.6\end{array}\) \(\begin{array}{lllllllll}4.6 & 3.9 & 3.9 & 5.3 & 2.8 & 2.6 & 3.0 & 3.3 & 3.7\end{array}\) Twenty-nine college students identified as having a negative attitude about Mitt Romney as compared to Barack Obama in the 2012 presidential election, were also asked to rate how trustworthy the face of Mitt Romney appeared. Here are the 29 ratings: \(\begin{array}{lllllllllll}1.8 & 3.3 & 4.3 & 4.4 & 2.5 & 2.6 & 3.5 & 4.2 & 4.7 & 2.5\end{array}\) \(\begin{array}{llllllllll}2.5 & 3.6 & 3.9 & 3.9 & 4.3 & 4.3 & 3.8 & 3.3 & 2.9 & 1.7\end{array}\) \(\begin{array}{lllllllll}3.3 & 3.3 & 3.9 & 4.3 & 4.1 & 3.8 & 3.3 & 5.3 & 5.4\end{array}\) (a) Do the sample means suggest that there is a difference in the mean trustworthy ratings between the two groups? (b) Make stemplots for both samples. Are there any obvious departures from Normality? (c) Test the hypothesis \(H_{0}: \mu_{1}=\mu_{2}\) against the one-sided alternative that students with a positive attitude rate Mitt Romney more trustworthy than those with a negative attitude. What do you conclude from part (a) and from the result of your test?

Short Answer

Expert verified
Part (a): Sample means suggest a difference. Part (b): Check stemplots for normality. Part (c): Hypothesis test to confirm difference.

Step by step solution

01

Calculate Sample Means for Each Group

For the group with a positive attitude, add all the ratings and divide by the number of students (29) to find the average. Do the same for the group with a negative attitude.
02

Compare Sample Means

Compare the two sample means you calculated. Determine if the mean of the group with a positive attitude is greater than the negative group, suggesting a possible difference.
03

Construct Stemplots for Each Group

Create a stemplot for each group to visualize the distribution of ratings. A stemplot helps identify any skewness or outliers in the data, assessing normality.
04

Evaluate Stemplots for Normality

Examine the stemplots to see if data is symmetrically distributed or if there are any significant departures from normality, such as skewness or outliers.
05

Set Up Hypotheses for the Test

State the null hypothesis: \(H_0: \mu_1 = \mu_2\) (means are equal). The alternative hypothesis is: \(H_a: \mu_1 > \mu_2\) (positive attitude ratings are higher).
06

Conduct Hypothesis Test

Use a t-test to compare the means of the two independent samples. Calculate the test statistic and find the p-value for the one-sided test.
07

Interpret Results of Hypothesis Test

If the p-value is less than the significance level (usually 0.05), reject the null hypothesis, indicating a significant difference in ratings. Otherwise, do not reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Means
In statistics, a sample mean is the average of all the data points collected in a sample. It serves as an estimate of the population mean, which represents the entire set from which the sample is drawn. In our exercise, we need to calculate the sample means for two groups of students. These groups provide ratings on how trustworthy Mitt Romney's face appeared during the 2012 election period. To compute the sample mean:
  • Add all the rating scores within each group.
  • Divide this sum by the number of ratings presented (in our case, 29 ratings per group).
After calculating, compare these means. Does the group with a positive attitude towards Mitt Romney rate him as more trustworthy than the group with a negative attitude? Observing such a difference might imply a variance in perception based on attitudes, but you will later need further statistical testing to substantiate this observation.
Stemplots
Stemplots, also known as stem-and-leaf plots, are a fantastic way to quickly visualize the distribution of a dataset. They're simple to construct and allow students to see the shape, central tendency, and spread of the data. In a stemplot, each number in the dataset is split into two parts: the stem, which is one of or more of the leading digits, and the leaf, which is usually the last digit. To create a stemplot:
  • Divide each rating number into a stem and a leaf.
  • List the stems in a vertical line.
  • Attach the leaves to their corresponding stems in increasing order.
This plot provides a clear view of the data, highlighting any peculiarities or trends. It’s also helpful in assessing the normality of the dataset. Through the stemplot, one can spot outliers or detect any skewness in the data distribution at a glance.
Normality Assessment
Normality assessment involves checking whether the data follows a normal distribution, which is a bell-shaped, symmetric distribution prevalent in nature and statistics. In observational studies, verifying normality is a crucial step because many statistical tests, like the t-test, assume that the data are normally distributed. When reviewing stemplots for normality, observe:
  • If the plot appears roughly symmetric around the center.
  • If the tails on both ends are evenly distributed.
  • For significant skews or outliers that could suggest deviations from normality.
Departure from normality can affect the results of hypothesis testing, potentially rendering them unreliable. Thus, before progressing with a t-test, ensure that any skewness is reasonable or apply necessary transformations to the data.
t-test
The t-test is a statistical test used to determine if there is a significant difference between the means of two groups. It's highly utilized in hypothesis testing to compare group means and assess whether observed differences are statistically meaningful or simply due to random chance. Here’s a simplified guide to conducting a t-test:
  • Establish your null hypothesis (\(H_0: \mu_1 = \mu_2\)) and alternate hypothesis (\(H_a: \mu_1 > \mu_2\)).
  • Calculate the t-statistic using the formula given for comparing means of two independent samples.
  • Find the p-value associated with the calculated t-statistic for the one-sided test.
  • Compare the p-value to the significance level (typically 0.05).
If the p-value is lower than the significance level, reject the null hypothesis, indicating a significant difference between groups. Otherwise, there isn’t enough evidence to conclude a difference exists. This process helps you derive insights about the data, confirming or challenging your initial observations.

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Most popular questions from this chapter

To study whether higher predictions are rated as more accurate than lower predictions, 161 college students were presented a prediction by a basketball expert about the winner of an upcoming basketball game between teams A and B. All participants learned that the expert had carefully examined the two teams' history, players, and other information. Eighty students were randomly assigned to a group in which the expert claimed the chance that A would win was \(70 \%\). The other 81 students were assigned to a group in which the expert claimed that the chance B would win was \(30 \%\). All students rated the expert on his accuracy using a scale ranging from 0 to 20 , with higher scores indicating greater accuracy. Are the conditions for two-sample \(t\) inference satisfied? (a) Maybe: the SRS condition is OK but we need to look at the data to check Normality. (b) No: scores in a range between 0 and 20 can't be Normal. (c) Yes: the SRS condition is OK and large sample sizes make the Normality condition unnecessary.

The data you used in the previous two problems came from a random sample of students who took the SAT twice. The response rate was \(63 \%\), which is pretty good for nongovernment surveys, so let's accept that the respondents do represent all students who took the exam twice. Nonetheless, we can't be sure that coaching actually caused the coached students to gain more than the uncoached students. Explain briefly but clearly why this is so.

Height and the big picture. Forty-six college students were randomly divided into two groups of size 23. One group was asked to imagine being on the upper floor of a tall building and the other on the lowest floor. Participants were then asked to choose between a job that required more detail orientation versus a job that required a more big-picture orientation. They rated their job preferences on an 11-point scale, with higher numbers corresponding to a greater preference for the big-picture job. Here are the summary statistics: 17 $$ \begin{array}{lccc} \hline \text { Group } & \text { Group Size } & \text { Mean } & \text { Standard Deviation } \\ \hline \text { Low } & 23 & 4.61 & 3.08 \\ \hline \text { High } & 23 & 6.68 & 3.45 \\ \hline \end{array} $$ (a) What degrees of freedom would you use in the conservative two-sample \(t\) procedures to compare the lower and higher floor groups? (b) What is the two-sample \(t\) test statistic for comparing the mean job preference ratings for the two groups? (c) Test the null hypothesis of no difference between the two population means against the two-sided alternative. Use your statistic from part (b) with degrees of freedom from part (a).

Researchers gave 40 index cards to a waitress at an Italian restaurant in New Jersey. Before delivering the bill to each customer, the waitress randomly selected a card and wrote on the bill the same message that was printed on the index card. Twenty of the cards had the message, "The weather is supposed to be really good tomorrow. I hope you enjoy the day!" Another 20 cards contained the message, "The weather is supposed to be not so good tomorrow. I hope you enjoy the day anyway!" After the customers left, the waitress recorded the amount of the tip (percent of bill) before taxes. Here are the tips for those receiving the good-weather message: \({ }^{20}\) \(\begin{array}{llllllllll}20.8 & 18.7 & 19.9 & 20.6 & 21.9 & 23.4 & 22.8 & 24.9 & 22.2 & 20.3\end{array}\) \(\begin{array}{llllllllll}24.9 & 22.3 & 27.0 & 20.5 & 22.2 & 24.0 & 21.2 & 22.1 & 22.0 & 22.7\end{array}\) The tips for the 20 customers who received the bad weather message are \(18.0 \quad 19.1 \quad 19.2 \quad 18.8 \quad 18.4 \quad 19.0 \quad 18.5 \quad 16.1 \quad 16.8 \quad 14.0\) \(\begin{array}{lllllllll}17.0 & 13.6 & 17.5 & 20.0 & 20.2 & 18.8 & 18.0 & 23.2 & 18.2\end{array}\) (a) Make stemplots or histograms of both sets of data. Because the distributions are reasonably symmetric with no extreme outliers, the \(t\) procedures will work well. (b) Is there good evidence that the two different messages produce different percent tips? State hypotheses, carry out a two-sample \(t\) test, and report your conclusions.

Healthy men aged 21-35 were randomly assigned to one of two groups: half received \(0.82\) gram of alcohol per kilogram of body weight; half received a placebo. Participants were then given 30 minutes to read up to 34 pages of Tolstoy's War and Peace (beginning at Chapter 1 , with each page containing approximately 22 lines of text). Every two to four minutes, participants were prompted to indicate whether they were "zoning out." The proportion of times participants indicated they were zoning out was recorded for each subject. The following table summarizes data on the proportion of episodes of zoning out. \({ }^{13}\) (The study report gave the standard error of the mean \(s / n \sqrt{n}\), abbreviated as SEM, rather than the standard deviation s.) $$ \begin{array}{llll} \hline \text { Group } & n & x^{-} \bar{x} & \text { SEM } \\ \hline \text { Alcohol } & 25 & 0.25 & 0.05 \\ \hline \text { Placebo } & 25 & 0.12 & 0.03 \\ \hline \end{array} $$ (a) What are the two sample standard deviations? (b) What degrees of freedom does the conservative Option 2 use for twosample \(t\) procedures for these samples? (c) Using Option 2, give a \(90 \%\) confidence interval for the mean difference between the two groups.

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