Question

Established 95% and 99% confidence interval about the mean for a psychological test that is given to thirty two persons and has a mean of 53 and a standard deviation of 7.5 assuming that the population standard deviation is not known.

Answer 1

1. 95% confidence interval 

The critical value for $\alpha = 0.05$ and $df = n-1 = 31$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.039513$

The corresponding confidence interval is computed as shown below:

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

$$=(53- 2.039513\times \dfrac{7.5}{\sqrt{32}}, 53+2.039513\times \dfrac{7.5}{\sqrt{32}})$$

$$=(50.296, 55.704)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $50.296 < \mu < 55.704,$ which indicates that we are 95% confident that the true population mean $\mu$

is contained by the interval $(50.296, 55.704).$

2. 99% confidence interval 

The critical value for $\alpha = 0.01$ and $df = n-1 = 31$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.744042$

The corresponding confidence interval is computed as shown below:

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

$$=(53- 2.744042\times \dfrac{7.5}{\sqrt{32}}, 53+2.744042\times \dfrac{7.5}{\sqrt{32}})$$

$$=(49.362, 56.638)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $49.362 < \mu < 56.638,$ which indicates that we are 99% confident that the true population mean $\mu$

is contained by the interval $(49.362, 56.638).$