Question

A random sample is drawn from a population of unknown standard deviation. Construct a 99% confidence interval for the population mean based on the information given:

a. N=49 ×=17.1 0=2.1

b. n=169 ×=17.1 0=2.1

Answer 1

a. The provided sample mean is $\bar{x}=17.1$  and the sample standard deviation is $s=2.1.$ The size of the sample is $n=49$ and the required confidence level is $99\%.$

The number of degrees of freedom are $df=n-1=49-1=48,$ and the significance level is $\alpha=0.01.$

Based on the provided information, the critical $t$ -value for $\alpha=0.01$ and $df=48$ degrees of freedom is $t_c=2.682204.$

The 99% confidence for the population mean $\mu$  is computed using the following expression

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

$$CI=(17.1-\dfrac{2.682204\times 2.1}{\sqrt{49}},17.1-\dfrac{2.682204\times 2.1}{\sqrt{49}} )$$

$$=(16.2953, 17.9047)$$

b. The provided sample mean is $\bar{x}=17.1$  and the sample standard deviation is $s=2.1.$ The size of the sample is $n=169$ and the required confidence level is $99\%.$

The number of degrees of freedom are $df=n-1=169-1=168,$ and the significance level is $\alpha=0.01.$

Based on the provided information, the critical $t$ -value for $\alpha=0.01$ and $df=168$ degrees of freedom is $t_c=2.60541.$

The 99% confidence for the population mean $\mu$  is computed using the following expression

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

$$CI=(17.1-\dfrac{2.60541\times 2.1}{\sqrt{169}},17.1-\dfrac{2.60541\times 2.1}{\sqrt{169}} )$$

$$=(16.6791, 17.5209)$$