Respuesta :
Answer:
99% confidence interval = [-35.47,-26.53]
Step-by-step explanation:
For Method 1:
[tex]n_1=66,\mu_1=54.9,\sigma_1=6.04[/tex]
For Method 2:
[tex]n_2=110,\mu_2=85.9,\sigma_2=16.44[/tex]
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.
[tex]Difference=\mu_1-\mu_2[/tex]
[tex]Difference=54.9-85.9[/tex]
[tex]Difference=-31.0[/tex]
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:
[tex]M.E.=z^*\times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}}[/tex]
Substitute the values.
[tex]M.E.=2.576\times \sqrt{\dfrac{(6.04)^2}{66}+\dfrac{(16.44)^2}{110}}}[/tex]
[tex]M.E.\approx 4.47[/tex]
Therefore, the margin of error is 4.47.
Step 3: Find 99% confidence interval.
[tex]C.I.=(\mu_1-\mu_2)\pm M.E.[/tex]
[tex]C.I.=-31.0\pm 4.47[/tex]
[tex]C.I.=-35.47,-26.53[/tex]
Therefore, the 99% confidence interval is [-35.47,-26.53].
