Question

A research firm conducted a survey to determine the mean amount of money smokers spend on cigarettes during a day. A sample of 100 smokers revealed that the sample mean is N$ 5.24 and sample standard deviation is N$ 2.18 Assume that the sample was drawn from a normal population.

2.1 Find the point estimate of the population mean.

2.2 Determine the lower limit of the 95% conference interval for estimating the unknown population

Answer 1

2.1. The sample mean $5.24 is a point estimate of the unknown population mean and is the best estimate, given the data.

2.2. The number of degrees of freedom are $df=100-1=99,$ and the significance level is $\alpha=0.05.$ The critical t-value for  $\alpha=0.05$ and $df=99$ degrees of freedom is $t_c=1.984217.$

The 95% confidence for the population mean $\mu$ is computed using the following expression

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

$$=(5.24-\dfrac{1.984217\times 2.18}{\sqrt{100}}, 5.24+\dfrac{1.984217\times 2.18}{\sqrt{100}})=$$

$$=(4.807, 5.673)$$

Lower limit is 4.807.