Question

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?

Answer 1

$$mean=\bar{x}=\dfrac{1}{6}(4+6+3+5+9+3)=5$$

$$s^2=\dfrac{1}{6-1}((4-5)^2+(6-5)^2+(3-5)^2+(5-5)^2$$

$$+(9-5)^2+(3-5)^2)=5.2$$

$$s=\sqrt{s^2}=\sqrt{5.2}\approx2.280351$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-2.015036\times\dfrac{2.280351}{\sqrt{6}},$$

$$5+2.015036\times\dfrac{2.280351}{\sqrt{6}})$$

$$=(3.1241, 6.8759)$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-2.570543\times\dfrac{2.280351}{\sqrt{6}},$$

$$5+2.570543\times\dfrac{2.280351}{\sqrt{6}})$$

$$=(2.6070, 7.3930)$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-4.031677\times\dfrac{2.280351}{\sqrt{6}},$$

$$5+4.031677\times\dfrac{2.280351}{\sqrt{6}})$$

$$=(1.2467, 8.7533)$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-1.710882\times\dfrac{2.280351}{\sqrt{25}},$$

$$5+1.710882\times\dfrac{2.280351}{\sqrt{25}})$$

$$=(4.2197, 5.7803)$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-2.063899\times\dfrac{2.280351}{\sqrt{25}},$$

$$5+2.063899\times\dfrac{2.280351}{\sqrt{25}})$$

$$=(4.0587, 5.9413)$$

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:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(5-2.79694\times\dfrac{2.280351}{\sqrt{25}},$$

$$5+2.79694\times\dfrac{2.280351}{\sqrt{25}})$$

$$=(3.7244, 6.2756)$$

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.