/*! 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 28 Student survey \(\quad\) Refer t... [FREE SOLUTION] | 91Ó°ÊÓ

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Student survey \(\quad\) Refer to the FL Student Survey data file on the text CD. Use the number of times reading a newspaper as the response variable and gender as the explanatory variable. The observations are as follows: $$ \begin{array}{ll} \text { Females: } & 5,3,6,3,7,1,1,3,0,4,7,2,2,7,3,0,5,0,4,4, \\ & 5,14,3,1,2,1,7,2,5,3,7 \\ \text { Males: } & 0,3,7,4,3,2,1,12,1,6,2,2,7,7,5,3,14,3,7, \\ & 6,5,5,2,3,5,5,2,3,3 \end{array} $$ Using software, a. Construct and interpret a plot comparing responses by females and males. b. Construct and interpret a \(95 \%\) confidence interval comparing population means for females and males. c. Show all five steps of a significance test comparing the population means. d. State and check the assumptions for part b and part \(c\).

Short Answer

Expert verified
The comparison shows whether gender influences newspaper reading frequency, with confidence intervals providing insights into mean differences, validated by significance testing.

Step by step solution

01

Organize the Data

The data is organized into two categories based on gender: Females and Males. For Females, the data points are 5, 3, 6, 3, 7, ... , 5, 3, 7. For Males, the data points are 0, 3, 7, 4, 3, ... , 3, 3.
02

Construct a Comparative Plot

Using software like R or Python, we can generate a boxplot or side-by-side histogram to visualize the distribution of newspaper reading frequency for both females and males. A boxplot will show median, quartiles, and potential outliers, allowing comparison of distributions at a glance.
03

Interpret the Plot

From the comparative plot, observe the center (median), spread (interquartile range), and any outliers in both distributions. This will help determine if one gender generally reads newspapers more frequently than the other, and if there's significant variability within genders.
04

Calculate the Mean and Standard Deviation

Compute the sample means and standard deviations for both females and males. The mean is calculated as the sum of all data points divided by the number of data points, and the standard deviation measures the dispersion from the mean.
05

Calculate the 95% Confidence Interval

Use the formula for a confidence interval for the difference of means: \[ \bar{x}_F - \bar{x}_M \pm t^* \sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}} \] where \( \bar{x}_F \) and \( \bar{x}_M \) are the sample means for females and males, \( s_F \) and \( s_M \) are the sample standard deviations, \( n_F \) and \( n_M \) are the sample sizes, and \( t^* \) is the critical t-value for the 95% confidence level.
06

Perform a Significance Test

Conduct a hypothesis test with the following steps: 1. **Null Hypothesis (H0):** There is no difference in population means (\( \mu_F = \mu_M \)). 2. **Alternative Hypothesis (H1):** There is a difference in population means (\( \mu_F eq \mu_M \)). 3. **Level of significance:** \( \alpha = 0.05 \). 4. **Test Statistic:** Compute the t-statistic using the formula: \[ t = \frac{(\bar{x}_F - \bar{x}_M)}{\sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}}} \]5. **Decision Rule:** Compare the calculated t-value to the critical t-value from the t-distribution table. Reject H0 if the absolute t-statistic is greater than the critical value.
07

State and Check Assumptions

Check assumptions for normality and equal variances using plots (e.g., QQ-plots) or statistical tests (Shapiro-Wilk for normality and F-test for equal variances). Assumptions include the samples being independent, drawn from normally distributed populations, and having approximately equal variances.

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Key Concepts

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

Boxplot
A boxplot is a visual tool that provides a quick overview of the distribution of data. It displays the minimum, first quartile, median, third quartile, and maximum of the dataset, which are collectively known as the five-number summary. This summary gives us essential insights about our data, such as center, spread, and potential outliers. When comparing two groups, like males and females in our newspaper reading example, a boxplot can illustrate differences in median reading frequency and variability.
The boxplot aids in identifying distribution symmetry and any gaps or unusual patterns. It visually depicts the interquartile range (IQR), which is the difference between the first (Q1) and third (Q3) quartiles, a measure of statistical dispersion. Outliers are shown as individual points that fall outside 1.5 times the IQR from Q1 and Q3.
For this exercise, a side-by-side boxplot helps us to quickly compare how often males versus females read newspapers, potentially revealing whether there are significant disparities or trends that might require further investigation.
Confidence Interval
Confidence intervals provide a range of values that likely contain a population parameter. For our example, we determine the 95% confidence interval to compare the means of newspaper reading between females and males. This confidence interval tells us that we are 95% confident the true difference in population means lies within this range.
Calculating a confidence interval involves determining the sample mean difference and the associated margin of error. The margin of error is derived from the standard error and the critical value from the t-distribution, reflecting the variability we can expect. In this case, the calculation formula is: \[\bar{x}_F - \bar{x}_M \pm t^* \sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}} \] This formula incorporates both the variability within columns and sample size, providing a robust measure of uncertainty around our estimate.
  • Interpretation: If a confidence interval for the difference excludes zero, it suggests a statistically significant difference in means. This interpretation allows students to understand whether gender influences newspaper reading frequency.
Significance Test
A significance test evaluates whether the observed differences between groups, like the newspaper reading habits of males and females, are likely due to chance. We use a t-test for this purpose, formulated under specific hypotheses.
A typical approach involves the following steps:
  • Null Hypothesis (H0): Assumes no difference in population means (bF = bM).
  • Alternative Hypothesis (H1): Assumes a difference exists (bF ≠ bM).
  • Test Statistic: The t-statistic is calculated to measure the difference relative to variability, using: \[t = \frac{(\bar{x}_F - \bar{x}_M)}{\sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}}} \]
  • Decision Rule: By comparing the t-statistic to the critical t-value from the t-distribution table at a 0.05 significance level, we decide whether to reject H0.
Assess the calculated p-value against the significance level (α = 0.05). If the p-value is smaller, it suggests that the sample data provide enough evidence to conclude a statistically significant difference in means.
Normality Assumption
Statistical tests, like the t-test used here, rely on certain assumptions. One key assumption is that the data for each group comes from normally distributed populations. This assumption ensures the test results are valid and reliable.
To check for normality, we utilize visual methods such as QQ-plots, or execute statistical tests like the Shapiro-Wilk test. If data points closely align with the diagonal line in a QQ-plot, the data may be considered normally distributed.
It's crucial to recognize that while minor deviations from normality might not severely impact the outcome, significant deviations could compromise test results.
  • Equal Variances: Another assumption often paired with normality is that both samples have equal variances. This is typically verified using an F-test. Ensuring both assumptions allows for accurate interpretation of any significance in the test results.
Understanding and verifying these assumptions equip students with the knowledge to effectively interpret the implications of their analyses.

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Most popular questions from this chapter

Female college student participation in athletics has increased dramatically over the past few decades. Sports medicine providers are aware of some unique health concerns of athletic women, including disordered eating. A study (M. Reinking and L. Alexander, Journal of Athletic Training, vol. \(40,2005, \mathrm{p}\). 47-51) compared disordered-eating symptoms and their causes for collegiate female athletes (in lean and nonlean sports) and nonathletes. The sample mean of the body dissatisfaction assessment score was \(13.2(s=8.0)\) for 16 lean sport athletes (those sports that place value on leanness, including distance running, swimming, and gymnastics) and \(7.3(s=6.0)\) for the 68 nonlean sport athletes. Assuming equal population standard deviations, a. Find the standard error for comparing the means. b. Construct a \(95 \%\) confidence interval for the difference between the mean body dissatisfaction for lean sport athletes and nonlean sport athletes. Interpret.

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