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% confidence interval for estimating the unknown population mean.

2.3) Determine the upper limit of the 95% confidence interval for estimating the unknown population mean.

Answer 1

2.1

the point estimate of the population mean is

`\mu = \= x = \$ 5.24`

2.2

The lower limit
`k= 4.8127`

2.3

The upper limit
`u= 5.6673`

Explanation:

From the question we are told that

The sample size is n = 100

The sample mean is
`\= x = \$ 5.24`

The standard deviation is
`\sigma = \$ 2.18`

Generally given that the sample size is large enough n > 30 and that the sample was drawn from a normal population, then the point estimate of the population mean is

`\mu = \= x = \$ 5.24`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

=>
`E = 1.96 *  (2.18  )/(√(100) )`

=>
`E =0.4273`

Generally the lower limit of the 95% confidence interval for estimating the unknown population mean is mathematically represented as

`k =  \= x  -E`

=>
`k= 5.24  -0.4273`

=>
`k= 4.8127`

Generally the upper limit of the 95% confidence interval for estimating the unknown population mean is mathematically represented as

`u=  \= x  + E`

=>
`u= 5.24  + 0.4273`

=>
`u= 5.6673`