Answer to Question #345161 in Statistics and Probability for Ekhang

Question

You are planning to buy a good quality cell phone in order to attend the online calss.The average price of 50 cellphones is ₱13,500 with a standard deviation of ₱573.65 and a confidence level of 99%. Find the corresponding confidence interval.

Answer 1

The critical value for $\alpha = 0.01,$ and $df = n-1 = 49$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =2.679952.$

The corresponding confidence interval is computed as shown below:

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

$$=(13500-2.679952\times\dfrac{573.65}{\sqrt{50}},$$

$$13500+2.679952\times\dfrac{573.65}{\sqrt{50}})$$

$$=(13282.585, 13717.415)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $13282.585 < \mu < 13717.415,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(13282.585, 13717.415).$