### Understanding the Difference in Pass Rates: Morning vs. Afternoon SectionsA university department chair believes that the afternoon section of a large survey class has a consistently lower proportion of students that pass the class compared to the morning section. To investigate this hypothesis, the chair checks a random sample of student records from each section. The results of the data collected are shown below.#### Data Summary:**Morning Section:**- Successes: 346- Observations: 400- Sample Proportion (\( \hat{p}_1 \)): 0.865**Afternoon Section:**- Successes: 426- Observations: 500- Sample Proportion (\( \hat{p}_2 \)): 0.852**Statistical Analysis:**- Confidence Level: 95%- \( z \)-score: -0.5548- \( p \)-value: 0.2912#### Interpretation of Results:The \( z \)-score and \( p \)-value indicate the statistical significance of the difference between the passing rates of the two sections. Given the \( p \)-value of 0.2912, which is higher than the common significance level of 0.05, we do not have sufficient evidence to reject the null hypothesis. This means there is no significant difference in the passing rates between the morning and afternoon sections at the 95% confidence level.#### Diagram Explanation:The provided figure is a normal distribution curve. The shaded area under the curve represents the region associated with the \( p \)-value. The curve extends from -3 to +3 standard deviations, with a marked \( z \)-score of -0.5548. Here, the \( z \)-score corresponds to the position on the curve relative to the mean, while the shaded area visually denotes the \( p \)-value region, showcasing that the result is not statistically significant.#### Calculation Details:To strengthen your understanding, you may manually input values and verify calculations:1. Morning section samples (\( n_1 \)): Example: 92. Afternoon section samples (\( n_2 \)): (Input specific values)3. Sample proportion for morning section (\( \hat{p}_1 \)): Example: 1.2344. Sample proportion for afternoon section (\( \hat{p}_2 \)): (Input specific values)By examining the data and statistical exhibits, we validate

Let n1 be the number of observations in the morning section. x1 be the number of successes in the…

A university department chair believes that the afternoon section of a large survey class has a consistently lower proportion of students that pass the class compared to the morning section. The chair checks a random sample of student records from each section. The results of the data collected are shown below. Morning section Afternoon section Successes 346 Successes 426 500 Observations 400 Observations p-hat_1 0.865 p-hat 2 0.852 Confidence level 95% z-score -0.5548 p-value 0.2912 -3 2 3 P = Ex: 1.234 Z= Morning section samples: n₁ = Ex: 9 Afternoon section samples: n₂ = Sample proportion for morning section samples: p₁ = Ex: 1.234 Sample proportion for afternoon section samples: P₂ =

A university department chair believes that the afternoon section of a large survey class has a consistently lower proportion of students that pass the class compared to the morning section. The chair checks a random sample of student records from each section. The results of the data collected are shown below. Morning section Afternoon section Successes 346 Successes 426 500 Observations 400 Observations p-hat_1 0.865 p-hat 2 0.852 Confidence level 95% z-score -0.5548 p-value 0.2912 -3 2 3 P = Ex: 1.234 Z= Morning section samples: n₁ = Ex: 9 Afternoon section samples: n₂ = Sample proportion for morning section samples: p₁ = Ex: 1.234 Sample proportion for afternoon section samples: P₂ =

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The percentage of Texans not covered by health care insurance in 2015 was 17%. The Texas Health and Human Services Commission (HHSC) has been charged with conducting a sample survey to obtain more current information. What sample size would you recommend if the HHSC’s goal is to estimate the current proportion of Texans without health care insurance with a margin of error of .03? Use a 95% confidence level. Repeat part (a) using a 99% confidence level.
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A college administrator is interested in determining the proportion of students who receive some sort of financial aid. Instead of examining the records for all students, the administrator randomly draws 120 students and finds that 48 of them are receiving financial aid. Use a 90% confidence interval to estimate the true proportion of students who receive financial aid.
The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey.Part of the survey gauges student optimism in landing a job after graduation. According to one year’s surveyresults, published in American Demographics, among the 1218 respondents, 733 said that they expecteddifficulty finding a job. Construct a 90% confidence interval for the proportion of students who expecteddifficulty finding a job. Interpret your result.
In a poll, 1100 adults in Nantucket were asked about their computer gaming habits. One finding was that 43% of respondents never computer game. Find and interpret a 95% confidence interval for the proportion of all adults in Nantucket who never play computer games
An internet study was mailed to 2336 subjects randomly. 1108 surveys were returned. Construct a 95% confidence interval for the proportion of the returned surveys. identify the margin of error E.
K Do freshmen spend significantly more time on social media than other class ranks? A recent study investigated the amount of time per day freshmen, sophomores, juniors, and seniors spent on social media while doing schoolwork. The students surveyed were U.S. residents from a 4-year, public, primarily residential institution. The accompanying computer output shows the mean time each cohort spent on social media and an ANOVA table. Construct a 95% confidence interval to compare the population mean time spent per day on social media between freshmen and seniors. Click the icon to view the summary statistics and ANOVA table. The 95% confidence interval is. (Round to one decimal place as needed.) View an ex NOV 9 Summary Statistics and ANOVA Table 6 n tv Mean Std. Dev. Source DF Class Error Total F6 FR 443 62.5 71.6 SS 99509 3 1562 7927979 1565 8027488 Print SO 330 56.5 68.5 MS 33170 5076 JU 383 72.0 80.9 Done 8 F 6.53 FB SE 410 50.4 62.9 P 0.000 ZA 9 F9 X k answer
Create an Excel spreadsheet. Conduct an independent sample t-test evaluating if cat owners or dog owners love their pet more. Ratings for cat owners- 7,6,5,9,5 Ratings for dog owners- 8,7,9,10 Write APA formatted results for thus analysis, including Cohen's d for the results and a bar graph with Error Bars showing 95 percent confidence intervals.
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give an example of a study for which the independent sample t-test is appropiate
A study examined the interactions in a children's museum. (Interactions by visitors to the museum included show-and-tell, learning, teaching, refocusing, participatory play, advocating, and disciplining interactions.) Researchers observed a sample of 154 meaningful interactions, of which 73 were led by children and 81 were led by adult caregivers the target point estimate is 0.474 what is The 90% confidence interval?