/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 Exercises 59 to 60 refer to the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.

Short Answer

Expert verified
(a) Mean ≈ 3.23, SD ≈ 1.11; (b) Adjusted Mean = 4.21, SD = 0.568; (c) Two-sample t-test may be appropriate with checks.

Step by step solution

01

Calculate Mean for 1 to 5 Scale

To find the mean for the 1 to 5 scale, multiply each rating by its frequency, sum these values, and divide by the total number of ratings.\[\text{Mean} = \frac{(1\times2) + (2\times3) + (3\times1) + (4\times13) + (5\times3)}{22} = \frac{71}{22} \approx 3.23\]
02

Calculate Variance for 1 to 5 Scale

First find the squared differences from the mean for each rating, multiply each by its frequency, sum these, and then divide by the number of ratings.\[\text{Variance} = \frac{(1-3.23)^2\times2 + (2-3.23)^2\times3 + (3-3.23)^2\times1 + (4-3.23)^2\times13 + (5-3.23)^2\times3}{22}\approx 1.226\]
03

Calculate Standard Deviation for 1 to 5 Scale

The standard deviation is the square root of the variance calculated in Step 2.\[\text{SD} = \sqrt{1.226} \approx 1.11\]
04

Adjust Mean for 0 to 4 Scale

Since the decision was to add 1 to each of the ratings from the 0 to 4 scale, the mean will also increase by 1. Thus, the adjusted mean is:\[\text{Adjusted Mean} = 3.21 + 1 = 4.21\]
05

Adjust Standard Deviation for 0 to 4 Scale

Adding a constant to each observation does not change the standard deviation, so the adjusted standard deviation remains the same as the original:\[\text{Adjusted SD} = 0.568\]
06

Evaluate Appropriateness of Two-Sample t-Test

The two sample t-test can be used when comparing means from independent samples, especially when the sample sizes are large enough to assume normality. Here, both samples have more than 20 observations, but we need to ensure the conditions of normality and equal variance are met.

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Key Concepts

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

Mean and Standard Deviation
Understanding the concepts of mean and standard deviation is crucial in statistics. The mean, often called the average, is the sum of all the data points divided by the number of data points. It's a measure of the central tendency of your data, meaning it represents the typical value. To calculate the mean of ratings on a 1 to 5 scale, you take each rating, multiply it by its frequency, and divide by the total number of ratings. This gives you the mean rating for that group, providing a single number that summarizes the entire set of data. The standard deviation, on the other hand, measures the amount of variation or dispersion in your data. A low standard deviation means the data points are close to the mean, while a high standard deviation indicates they are spread out over a wider range. You find the standard deviation by first calculating the variance, which involves finding the squared differences from the mean, multiplying by their frequencies, and then averaging these values. The standard deviation is the square root of this variance.
Two-Sample t-Test
The two-sample t-test is a method used to determine if two populations have different means. This test helps decide whether the difference in means between two samples is significant. For the t-test to be appropriate, several conditions need to be met:
  • Independence: The samples must be independent of each other.
  • Normality: The data should be approximately normally distributed, or the sample sizes should be large enough (usually n > 30) to rely on the Central Limit Theorem.
  • Equal variances: The variance within each sample should be roughly equal, although this isn't strictly required with some variations of the test.
In the exercise, these conditions were evaluated to decide if the test could be applied to compare the means of the two groups surveyed with different scales. This involved considering the sample sizes and distributions to ensure the method was suitable.
Sampling Methods
Sampling methods are techniques used to select a group of subjects or items from a larger population. The goal is to obtain a sample that is representative of the larger group. One popular method is Simple Random Sampling (SRS), where every member of the population has an equal chance of being selected. This method helps to avoid bias in your sample results. For example, selecting students alphabetically from a roster ensures that each student has an equal opportunity to be chosen, leading to results that more accurately reflect the population as a whole. Understanding sampling methods is vital in statistics because the way a sample is collected can significantly affect the validity of the conclusions drawn from it. Proper sampling techniques help ensure that the results of your analysis are reliable and can be generalized to the entire population.
Survey Design
Survey design involves creating a set of questions to gather data from respondents. A well-designed survey should yield clear, unbiased results that answer the research question. Key factors to consider in survey design include:
  • Question wording: Questions should be neutral and easy to understand, avoiding leading or loaded phrasing.
  • Scale: Choosing the right scale for responses is crucial, as the exercise showed with the 1 to 5 vs. 0 to 4 scales. Different scales can influence how respondents perceive and answer questions.
  • Sample size: The number of people surveyed should be enough to provide statistically significant results, yet manageable within logistical constraints.
By carefully tailoring a survey design, researchers can effectively study population characteristics and achieve a deeper understanding of the data gathered.

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Most popular questions from this chapter

Students and catalog shopping What is the most important reason that students buy from catalogs? The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large midwestern university: \(^{26}\) $$ \begin{array}{lcc} \hline & \text { American } & \text { Asian } \\ \text { Save time } & 29 & 10 \\ \text { Easy } & 28 & 11 \\ \text { Low price } & 17 & 34 \\ \text { Live far from stores } & 11 & 4 \\ \text { No pressure to buy } & 10 & 3 \\ \hline \end{array} $$ (a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). (c) Check that the conditions for carrying out the test are met. (d) Interpret the \(P\) -value in context. What conclusion would you draw?

Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ The appropriate null hypothesis for performing a chi-square test is that (a) equal proportions of female and male teenagers are almost certain they will be married in 10 years. (b) there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers' opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population.

Refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ (a) \(\frac{(18-25)^{2}}{25}+\frac{(22-25)^{2}}{25}+\frac{(39-25)^{2}}{25}+\frac{(21-25)^{2}}{25}\) (b) \(\frac{(25-18)^{2}}{18}+\frac{(25-22)^{2}}{22}+\frac{(25-39)^{2}}{39}+\frac{(25-21)^{2}}{21}\) (c) \(\frac{(18-25)}{25}+\frac{(22-25)}{25}+\frac{(39-25)}{25}+\frac{(21-25)}{25}\) (d) \(\frac{(18-25)^{2}}{100}+\frac{(22-25)^{2}}{100}+\frac{(39-25)^{2}}{100}+\frac{(21-25)^{2}}{100}\) (e) \(\frac{(0.18-0.25)^{2}}{0.25}+\frac{(0.22-0.25)^{2}}{0.25}+\frac{(0.39-0.25)^{2}}{0.25}\) \(+\frac{(0.21-0.25)^{2}}{0.25}\) The chi-square statistic is

Multiple choice: Select the best answer for Exercises 19 to 22 Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, "Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?" Here are her data: $$ \begin{array}{lcccc} \hline \text { Type of Food: } & \text { Asian } & \text { Mexican } & \text { Pizza } & \text { Hamburgers } \\ \text { Count: } & 18 & 22 & 39 & 21 \\ \hline \end{array} $$ An appropriate null hypothesis to test whether the food choices are equally popular is (a) \(H_{0}: \mu=25,\) where \(\mu=\) the mean number of students that prefer each type of food. (b) \(H_{0}: p=0.25,\) where \(p=\) the proportion of all students who prefer Asian food. (c) \(H_{0}: n_{A}=n_{M}=n_{P}=n_{H}=25,\) where \(n_{A}\) is the number of students in the school who would choose Asian food, and so on. (d) \(H_{0}: p_{A}=p_{M}=p_{P}=p_{H}=0.25,\) where \(p_{A}\) is the proportion of students in the school who would choose Asian food, and so on. (e) \(\quad H_{0}: \hat{p}_{\mathrm{A}}=\hat{p}_{M}=\hat{p}_{P}=\hat{p}_{H}=0.25,\) where \(\hat{p}_{\mathrm{A}}\) is the pro- portion of students in the sample who chose Asian food, and so on.

Skittles Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: "The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor." (a) State appropriate hypotheses for a significance test of the company's claim. (b) Find the expected counts for a bag of Skittles with 60 candies. (c) How large a \(\chi^{2}\) statistic would you need to have significant evidence against the company's claim at the \(\alpha=0.05\) level? At the \(\alpha=0.01\) level? (d) Create a set of observed counts for a bag with 60 candies that gives a \(P\) -value between 0.01 and \(0.05 .\) Show the calculation of your chi-square statistic.

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