Question

A researcher compares the effectiveness of two different instructional methods for teaching anatomy. A sample of 215 students using Method 1 produces a testing average of 55.5. A sample of 242 students using Method 2 produces a testing average of 64.2. Assume that the population standard deviation for Method 1 is 7.95, while the population standard deviation for Method 2 is 18.21. Determine the 90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 2 of 3 : Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

Answer #1

- Here let X1 be the random variable denoting the effectiveness of the teaching instructional method 1
- Here let X2 be the random variable denoting the effectiveness of the teaching instructional method 2
- to calculate the 90% confidence interval
- 0.90=P(-1.645<N(0,1)<1.645)
- 0.90=P(-1.645<<1.645)
- substituing the above value for mean ,standard deviation and n
- The confidence interval obtained is as follows.
- (-10.8221<<-6.577)
- This shows that effectiveness of instruction of method2 is more than method 1.
- b)To calculate the margin of error
- margin of error is calculated as critical value *standard error
- =1.645*
- =1.645*1.29=2.122132

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