Question

Assuming that the samples come from normal distributions, find the margin of error E given the following.

2. n=150, x-=98, s=9.90, 95% confidence.

3. n=350, x-=125, s=11.18, 99% confidence.

4. n=500, x-=236, s=15.36, 90% confidence.

5. n=785, x-=459, s=21.42, 99% confidence.

Answer 1

2. The critical value for $\alpha = 0.05,df=n-1=149$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 1.976013.$

Margin of Error: $E=t_c\times\dfrac{s}{\sqrt{n}}=1.976013 \times \dfrac{9.9}{\sqrt{150}}= 1.5973$

3. The critical value for $\alpha = 0.01,df=n-1=349$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.58999.$

Margin of Error: $E=t_c\times\dfrac{s}{\sqrt{n}}=2.58999\times \dfrac{11.18}{\sqrt{350}}= 1.5478$

4. The critical value for $\alpha = 0.10,df=n-1=499$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 1.647913.$

Margin of Error: $E=t_c\times\dfrac{s}{\sqrt{n}}=1.647913 \times \dfrac{15.36}{\sqrt{500}}=1.1320$

5. The critical value for $\alpha = 0.01,df=n-1=784$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.582115.$

Margin of Error: $E=t_c\times\dfrac{s}{\sqrt{n}}=2.582115 \times \dfrac{21.42}{\sqrt{785}}= 1.9741$