91Ó°ÊÓ

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: \({ }^{19}\) \(\begin{array}{llllllllll}2.6 & 3.2 & 3.7 & 3.3 & 3.4 & 3.6 & 3.7 & 3.8 & 3.9 & 4.1\end{array}\) \(\begin{array}{lllllllllll}4.2 & 4.9 & 5.7 & 4.2 & 3.9 & 3.2 & 4.5 & 5.0 & 5.0 & 4.6\end{array}\) \(\begin{array}{lllllllll}4.6 & 3.9 & 3.9 & 5.3 & 2.8 & 2.6 & 3.0 & 3.3 & 3.7\end{array}\) (a) Suppose we can consider this an SRS of all U.S. college students. Make a stemplot. Is there any sign of major deviation from Normality? (b) Give a \(95 \%\) confidence interval for the mean rating. (c) Is there significant evidence at the \(5 \%\) level that the mean rating is greater than \(3.5\) (a neutral rating)?

Step by step solution

01

Arrange Data for Stemplot

To create a stemplot, we'll break each data point into a 'stem' (the integer part) and a 'leaf' (the first decimal part). For instance, the number 4.6 would have a stem of 4 and a leaf of 6. All the numbers should be organized into this format.

02

Create the Stemplot

The stemplot shows the frequency of ratings by listing the stem on the left and the leaves on the right:\[ \begin{array}{c|l}2 & 6, 6, 8 \3 & 0, 2, 2, 3, 3, 4, 6, 7, 7, 7, 8, 9, 9, 9 \4 & 1, 2, 2, 5, 6, 6, 9 \5 & 0, 0, 3, 7 \\end{array} \] The stemplot shows a roughly symmetric distribution, indicating no major deviation from normality.

03

Calculate the Mean and Standard Deviation

First, calculate the mean (average) of all ratings: \( \bar{x} = \frac{\Sigma x}{n} \), where \( \Sigma x = 114.9\) and \(n = 29\). \( \bar{x} = \frac{114.9}{29} \approx 3.96 \). Next, find the standard deviation using \( s = \sqrt{\frac{\Sigma (x_i - \bar{x})^2}{n - 1}} \), which is approximately 0.84.

04

Calculate 95% Confidence Interval

Use the formula for the confidence interval: \( \bar{x} \pm t^* \frac{s}{\sqrt{n}} \), where \(t^*\) is the t-value for \(n-1 = 28\) degrees of freedom at 95% confidence, which is approximately 2.048. The interval is \( 3.96 \pm 2.048 \times \frac{0.84}{\sqrt{29}} \), resulting in (3.62, 4.30).

05

Perform Hypothesis Testing for Mean Greater Than 3.5

Set up the hypothesis: \(H_0: \mu = 3.5\) and \(H_a: \mu > 3.5\). Calculate the t-statistic: \( t = \frac{\bar{x} - 3.5}{s/\sqrt{n}} = \frac{3.96 - 3.5}{0.84/\sqrt{29}} \approx 2.98 \). Compare \(t\) with the critical value from the t-distribution. Since \(t = 2.98 > 2.048\), the result is significant at the 5% level, and we reject \(H_0\).

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values that estimate a population parameter. Here, we want to find out the true mean trustworthiness rating for all U.S. college students who have positive attitudes toward Mitt Romney. This estimate is not precise; it comes with some degree of uncertainty.

To express this uncertainty, we use a confidence interval. Typically, confidence intervals are calculated using a specified confidence level, like 95% in our exercise. This level tells us how sure we are that the interval contains the true population mean. In our exercise, the confidence interval for the mean trustworthiness rating was found to be between 3.62 and 4.30.

The calculation of this confidence interval involved first finding the mean of all ratings and then determining the standard deviation. With these, we can apply the formula:

• 
  
• 
  We add and subtract this margin from the mean to find the interval limits.

Hypothesis Testing

Hypothesis testing is a statistical method used to determine if there is sufficient evidence to support a specific hypothesis about a population parameter. In this exercise, we are testing if the mean rating is greater than 3.5.

First, we set up our hypotheses:

• The null hypothesis ( ) states there is no effect or difference, so here
• The alternative hypothesis ( ) claims a difference or effect exists, that is
  To test this, we calculate a t-statistic using the formula: This statistic tells us how far our sample mean (
  We then compare this t-statistic with a critical value from the t-distribution table. If our statistic is larger than the critical value at the 5% significance level, we reject the null hypothesis, which is what happened here. Consequently, we have significant evidence to say the mean rating is greater than 3.5.

Normality

Normality refers to how data is distributed. In statistics, many tests assume that the data distribution follows a bell curve, or normal distribution. Ensuring this assumption is met is crucial because it allows for more accurate and reliable results.

In this exercise, we assessed normality using a stemplot. A stemplot is a visual tool that displays data distribution. When the data points are symmetric along the plot, it suggests normality. The ratings we examined showed a roughly symmetric pattern, indicating no major deviation from normality.

Why is normality important? When data is normally distributed, statistical measures like the mean, variance, and distribution are more predictable. It also smoothes out the influence of outliers, making hypothesis testing results more robust.

• Normality allows us to use many statistical models and tests
• Ensures interpretation of parameters is more accurate

Stemplot

A stemplot, or stem-and-leaf plot, is a handy tool for visualizing numerical data. It helps us see the shape of the distribution and spot any patterns at a glance. This simple, yet effective tool is made by splitting each number into a stem (the leading digit(s)) and a leaf (the trailing digit).

For our exercise, the data was split into stems and leaves to reveal a distribution pattern. Stems represented whole number parts of ratings, while leaves indicated the decimal parts. For example, a rating of 4.6 had a stem of 4 and a leaf of 6. The stemplot showed how frequently each rating or range appeared.

Advantages of using a stemplot include:

• Maintains the original data values while displaying them
• Quickly reveals the shape of data distribution, such as symmetry or skewness

The stemplot in our task indicated a symmetric distribution, suggesting that the data was likely normally distributed, simplifying subsequent statistical analysis.