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

Answer:

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  $

Step-by-step 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  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }$

=>   $E = 1.96 *  \frac{2.18  }{\sqrt{100} }$

=>   $E =0.4273$

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  $