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.
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"
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