Find a 95% confidence interval for the difference between full attendance students' Exam 5 score and less attendance students' Exam 5 score. Assume equal population variance. The data are given below: Less attendance: 90, 90, 60, 50, 69, 88, 68, 85, 65Full attendance: 96, 96, 100, 99 Now find the margin of error. Round to 2 decimal places. **Problem Overview: Confidence Interval Calculation**This exercise involves calculating a 95% confidence interval for the difference in Exam 5 scores between two groups of students based on their attendance. It is assumed that the population variance is equal. The data provided is as follows:- **Less Attendance Group Scores:** 90, 90, 60, 50, 69, 88, 68, 85, 65- **Full Attendance Group Scores:** 96, 96, 100, 99**Instructions:**1. **Part 1: Margin of Error** - Calculate the margin of error for the difference in scores and round the result to two decimal places.2. **Part 2: Lower Limit of the Confidence Interval (CI)** - Determine the lower limit of the 95% confidence interval and round the result to two decimal places.3. **Part 3: Upper Limit of the Confidence Interval (CI)** - Calculate the upper limit of the 95% confidence interval and round the result to two decimal places.**Interface Features:**The interface includes interactive elements to input calculated values and navigate through the steps.

We have given data, Less attendance Full attendance 90 96 90 96 60 100 50 99 69 88 68 85 65 From the given data set, we used to do further calculation Less attendance xi Full attendance yi xi2 yi2 90 96 8100 9216 90 96 8100 9216 60 100 3600 10000 50 99 2500 9801 69 4761 88 7744 68 4624 85 7225 65 4225 Total 665 391 50879 38233 n1=9, n2=4, ∑xi=665, ∑yi=391, ∑xi2=50879 and ∑ yi2=38233 x¯=∑ xin1=6659=73.8889 y¯=∑ yin=3914=97.75 s12=1n1-1∑ xi2-n x¯2=19-150879-9×73.88892=18(1142.874111)≅217.8593 s22=1n2-1∑ yi2-n y¯2=14-138233-4× 97.752=1338233-4× 97.752=12.753=4.25 The sample pooled variance sp2=(n1-1) s12 +(n2-1) s22n1+n2-2=9-1×217.8593+4-1×4.259+4-2=8×217.8593+3×4.2511=1755.624411≅159.60221 sp=159.60221≅12.6336 Here the population standard deviation is unknown and have equal variance, so the random variable follows approximately t-distribution. The critical value of two-tailed t-test statistic for two sample test is t(n1+n2-2,α/2) The degrees of freedom are obtained as df=n1+n2-2=9+4-2=11 The level of significance is 0.05. The critical value for the two-tailed t-test is obtained using Excel as follows Click on any blank cell in Excel and type the below formula =T.INV.2T(0.05,11) The approximate critical value of the t-test is t(11,0.025)=2.2010. Part 1) The margin of error is ME=t(n1+n2-2,α/2).sp1n1+1n2=2.2010×12.633619+14=2.2010×12.6336×0.6009≅16.71 Part2) The lower limit of 95% confidence interval is obtained as Lower Limit=y¯-x¯-ME =97.75-73.8889-16.71≅7.15 Therefore the lower limit of 95% confidence interval is 7.15.Part 3) The upper limit of 95% confidence interval is obtained as Upper Limit=y¯-x¯+ME =97.75-73.8889+16.71≅40.57 Therefore the upper limit of 95% confidence interval is 40.57.

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Find a 95% confidence interval for the difference between full attendance students' Exam 5 score and less attendance students' Exam 5 score. Assume equal population variance. The data are given below: Less attendance: 90, 90, 60, 50, 69, 88, 68, 85, 65 Full attendance: 96, 96, 100, 99 Now find the margin of error. Round to 2 decimal places. Part 2 of 3 Olo Part 3 of 3 X Now find the lower limit of the 95% CI. Round to 2 decimal places. Olo X 2/6 □ 0/0 2:16 Now find the upper limit of the 95% CI. Round to 2 decimal places. -

Find a 95% confidence interval for the difference between full attendance students' Exam 5 score and less attendance students' Exam 5 score. Assume equal population variance. The data are given below: Less attendance: 90, 90, 60, 50, 69, 88, 68, 85, 65 Full attendance: 96, 96, 100, 99 Now find the margin of error. Round to 2 decimal places. Part 2 of 3 Olo Part 3 of 3 X Now find the lower limit of the 95% CI. Round to 2 decimal places. Olo X 2/6 □ 0/0 2:16 Now find the upper limit of the 95% CI. Round to 2 decimal places. -

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Question

Find a 95% confidence interval for the difference between full attendance students' Exam 5 score and less attendance students' Exam 5 score. Assume equal population variance.

The data are given below:

Less attendance: 90, 90, 60, 50, 69, 88, 68, 85, 65

Full attendance: 96, 96, 100, 99

Now find the margin of error. Round to 2 decimal places.

**Problem Overview: Confidence Interval Calculation**

This exercise involves calculating a 95% confidence interval for the difference in Exam 5 scores between two groups of students based on their attendance. It is assumed that the population variance is equal. The data provided is as follows:

- **Less Attendance Group Scores:** 90, 90, 60, 50, 69, 88, 68, 85, 65
- **Full Attendance Group Scores:** 96, 96, 100, 99

**Instructions:**

1. **Part 1: Margin of Error**
- Calculate the margin of error for the difference in scores and round the result to two decimal places.

2. **Part 2: Lower Limit of the Confidence Interval (CI)**
- Determine the lower limit of the 95% confidence interval and round the result to two decimal places.

3. **Part 3: Upper Limit of the Confidence Interval (CI)**
- Calculate the upper limit of the 95% confidence interval and round the result to two decimal places.

**Interface Features:**
The interface includes interactive elements to input calculated values and navigate through the steps.

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