Question

Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. the design of the study justifies the assumption that the sample can be treated as a simple random sample. use the data to construct a 90% confidence interval estimate of μ, the mean duration of game play.
4041, 3894, 3865, 4037, 4316, 4803, 4660, 4027, 5000, 4821, 4334, 4311

what is the confidence interval estimate of the population mean μ?

Answer 1

Answer:

The confidence interval estimate of the population mean = (4139.28,4545.56)

Step-by-step explanation:

We are given the following data set:

4041, 3894, 3865, 4037, 4316, 4803, 4660, 4027, 5000, 4821, 4334, 4311

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{52109}{12} = 4342.42$

Sum of squares of differences = 90852.00696 + 201077.507 + 227926.6736 + 93279.3403 + 697.840278 + 212137.0069 + 100859.1736, +99487.67363 + 432415.8402 + 229042.0069 + 70.84027778 + 987.0069447 = 1688832.917

$S.D = \sqrt{\frac{1688832.917}{11}} = 391.83$

Confidence interval:

$\bar{x} \pm t_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 11 and at}~\alpha_{0.10} = \pm 1.795885$

$4342.42 \pm 1.795885(\frac{391.83}{\sqrt{12}} ) = 4342.42 \pm 203.14 = (4139.28,4545.56)$