Aresearcher compares the effectiveness of two different instructional methods for teaching anatomy. a sample of 66 students using method 1 produces a testing average of 54.9. a sample of 110 students using method 2 produces a testing average of 85.9. assume that the population standard deviation for method 1 is 6.04, while the population standard deviation for method 2 is 16.44. determine the 99% confidence interval for the true difference between testing averages for students using method 1 and students using method 2. step 1 of 3 : find the point estimate for the true difference between the population means.

99% confidence interval = [-35.47,-26.53]

Step-by-step explanation:

For Method 1:

For Method 2:

We need to find the 99% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 1: Find the point estimate for the true difference between the population means.

Therefore the point estimate for the true difference between the population means is -31.

Step 2: Find margin of error at 99% confidence.

From the standard normal table the critical value of z at 99% confidence = 2.576.

The formula for margin of error:

Substitute the values.

Therefore, the margin of error is 4.47.

Step 3: Find 99% confidence interval.

Therefore, the 99% confidence interval is [-35.47,-26.53].

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15