The following simple random sample was selected from a normal distribution: 4, 6, 3, 5, 9, 3.
1. Construct a 90% confidence interval for the population mean μ.
2. Construct a 95% confidence interval for the population mean μ.
3. Construct a 99% confidence interval for the population mean μ.
4. Assume that the sample mean x and sample standard deviation s remain exactly the same as those you just calculated but that they are based on a sample of n = 25 observations rather than n = 6 observations. Repeat parts 1-3. What is the effect of increasing the sample size on the width of the confidence intervals?
"s^2=\\dfrac{1}{6-1}((4-5)^2+(6-5)^2+(3-5)^2+(5-5)^2"
"+(9-5)^2+(3-5)^2)=5.2"
1. The critical value for "\\alpha = 0.1" and "df = n-1 = 5" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =2.015036." The corresponding confidence interval is computed as shown below:
"=(5-2.015036\\times\\dfrac{2.280351}{\\sqrt{6}},"
"5+2.015036\\times\\dfrac{2.280351}{\\sqrt{6}})"
"=(3.1241, 6.8759)"
Therefore, based on the data provided, the 90% confidence interval for the population mean is "3.1241 < \\mu < 6.8759," which indicates that we are 90% confident that the true population mean "\\mu" is contained by the interval "(3.1241, 6.8759)."
2. The critical value for "\\alpha = 0.05" and "df = n-1 = 5" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =2.570543." The corresponding confidence interval is computed as shown below:
"=(5-2.570543\\times\\dfrac{2.280351}{\\sqrt{6}},"
"5+2.570543\\times\\dfrac{2.280351}{\\sqrt{6}})"
"=(2.6070, 7.3930)"
Therefore, based on the data provided, the 95% confidence interval for the population mean is "2.6070 < \\mu < 7.3930," which indicates that we are 95% confident that the true population mean "\\mu" is contained by the interval "(2.6070, 7.3930)."
3. The critical value for "\\alpha = 0.01" and "df = n-1 = 5" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =4.031677." The corresponding confidence interval is computed as shown below:
"=(5-4.031677\\times\\dfrac{2.280351}{\\sqrt{6}},"
"5+4.031677\\times\\dfrac{2.280351}{\\sqrt{6}})"
"=(1.2467, 8.7533)"
Therefore, based on the data provided, the 99% confidence interval for the population mean is "1.2467 < \\mu < 8.7533," which indicates that we are 99% confident that the true population mean "\\mu" is contained by the interval "(1.2467, 8.7533)."
4.
a. The critical value for "\\alpha = 0.1" and "df = n-1 = 24" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =1.710882." The corresponding confidence interval is computed as shown below:
"=(5-1.710882\\times\\dfrac{2.280351}{\\sqrt{25}},"
"5+1.710882\\times\\dfrac{2.280351}{\\sqrt{25}})"
"=(4.2197, 5.7803)"
Therefore, based on the data provided, the 90% confidence interval for the population mean is "4.2197 < \\mu < 5.7803," which indicates that we are 90% confident that the true population mean "\\mu" is contained by the interval "(4.2197, 5.7803)."
b. The critical value for "\\alpha = 0.05" and "df = n-1 = 24" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =2.063899." The corresponding confidence interval is computed as shown below:
"=(5-2.063899\\times\\dfrac{2.280351}{\\sqrt{25}},"
"5+2.063899\\times\\dfrac{2.280351}{\\sqrt{25}})"
"=(4.0587, 5.9413)"
Therefore, based on the data provided, the 95% confidence interval for the population mean is "4.0587 < \\mu < 5.9413," which indicates that we are 95% confident that the true population mean "\\mu" is contained by the interval "(4.0587, 5.9413)."
c. The critical value for "\\alpha = 0.01" and "df = n-1 = 24" degrees of freedom is "t_c = z_{1-\\alpha\/2; n-1} =2.79694." The corresponding confidence interval is computed as shown below:
"=(5-2.79694\\times\\dfrac{2.280351}{\\sqrt{25}},"
"5+2.79694\\times\\dfrac{2.280351}{\\sqrt{25}})"
"=(3.7244, 6.2756)"
Therefore, based on the data provided, the 99% confidence interval for the population mean is "3.7244 < \\mu < 6.2756," which indicates that we are 99% confident that the true population mean "\\mu" is contained by the interval "(3.7244, 6.2756)."
The width of a confidence interval decreases as the sample size increases and increases as the confidence level increases.
Larger samples give narrower intervals. We are able to estimate a population mean more precisely with a larger sample size.
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