Question

A simple random sample of 10 households, the number of TV's that each household had is as follows: 2 , 0 , 2 , 2 , 2 , 2 , 1 , 5 , 3 , 2 Assume that it is reasonable to believe that the population is approximately normal and the population standard deviation is 0.55 . What is the lower bound of the 95% confidence interval for the mean number of TV's?

Answer 1

Answer: 1.758 is the lower bound of the 95% confidence interval for the mean number of TV's.

Step-by-step explanation:

Given that,

n = 10

Number of TV each household have = {2 , 0 , 2 , 2 , 2 , 2 , 1 , 5 , 3 , 2}

Standard Deviation(SD) = 0.55

95% Confidence Interval,  = 0.05

Follows normal distribution,

Mean = $\bar{X} = \frac{2+0+2+2+2+2+1+5+3+2}{10}$

= $\frac{21}{10}$

= 2.1

Therefore, 95% Confidence Interval are as follows:

$\bar{X}\pm Z_{\frac{\alpha}{2}} \times \frac{\sigma}{\sqrt{n}}$

$2.1\pm 1.96 \times \frac{\0.55}{\sqrt{10}}$

Hence,

Lower bound = 2.1- 1.96 ×  $\frac{\0.55}{\sqrt{10}}$

                      = 2.1- 1.96 × 0.174

                      = 1.758