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.

2.1

  the point estimate of the population mean is

2.2

The lower limit          

2.3

The upper limit          

Step-by-step explanation:

From the question we are told that

   The sample size is  n  = 100

   The sample mean is  

    The standard deviation is  

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

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

=>  

Generally from the normal distribution table the critical value  of   is  

Generally the margin of error is mathematically represented as  

=>  

=>  

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

=>  

=>  

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

=>  

=>  

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: