Question

The data below refers to the time in hours spent on mobile internet by sample of 10 students in a class. 39 42 47 45 32 45 37 34 33 29 Assume that the population data follows a normal distribution with unknown mean and unknown standard deviation. Find a 95% confidence interval estimate of μ . [33.813, 42.787] hours [35.557, 41.936] hours [34.664, 41.936] hours

Answer 1

Answer:

Option A) (33.813,42.787)

Step-by-step explanation:

We are given the following data set:

39, 42, 47, 45, 32, 45, 37, 34, 33, 29

Sample size, n = 10

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.  

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

$Mean =\displaystyle\frac{383}{10} = 38.3$

Sum of squares of differences = 354.1

$S.D = \sqrt{\frac{354.1}{9}} = 6.27$

95% Confidence interval:  

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

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.26$  

$38.3 \pm 2.262(\frac{6.27}{\sqrt{10}} ) = 38.3\pm 4.484= (33.813,42.787)$