Mathematics
09.03.2022 21:24
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Aresearcher compares the effectiveness of two different instructional methods for

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.
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sakugrey
sakugrey
4,4(75 marks)

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

Step-by-step explanation:

For Method 1:

n_1=66,\mu_1=54.9,\sigma_1=6.04

For Method 2:

n_2=110,\mu_2=85.9,\sigma_2=16.44

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.

Difference=\mu_1-\mu_2

Difference=54.9-85.9

Difference=-31.0

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:

M.E.=z^*\times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}

Substitute the values.

M.E.=2.576\times \sqrt{\dfrac{(6.04)^2}{66}+\dfrac{(16.44)^2}{110}}}

M.E.\approx 4.47

Therefore, the margin of error is 4.47.

Step 3: Find 99% confidence interval.

C.I.=(\mu_1-\mu_2)\pm M.E.

C.I.=-31.0\pm 4.47

C.I.=-35.47,-26.53

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

mercedesamatap21hx0
mercedesamatap21hx0
4,6(97 marks)

7 or more hours

15 + 15 is 30

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