Answer to Question #142528 in Statistics and Probability for Ray Rietz

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)

2.2) Determine the lower limit of the 95% confidence interval for estimating the unknown population mean. (5)

2.3) Determine the upper limit of the 95% confidence interval for estimating the unknown population mean. (3)

2.4) Find the width of the 95% confidence interval for estimating the unknown population mean (2)

2.5) Use the data provided above to test a claim at a 1 % Level of Significance that the unknown population mean of spending amount on cigarettes is N$ 5.00 (8)

Answer 1

2.1)

The point estimate

$\mu = \bar x = 5.24$

2.2)

At 95% confidence interval, Z = -1.96 for lower limit

$$Z =\frac{X-\bar X}{\sigma}$$

$$X= Z\sigma + \bar X$$

$$X = -1.96 * 2.18 + 5.24 = 0.9672$$

2.3)

At 95% confidence interval, Z = 1.96 for lower limit

$X = 1.96 * 2.18 + 5.24 = 9.5128$

2.4)

$9.5128-0.9672= 8.5456$

2.5)

$$Z =\frac{X-\bar X}{\sigma}$$

$$Z =\frac{5-5.24}{2.18}= -0.1101$$

The 2 tail probability given Z = -0.1101 is 0.912409.

At a 1% level of significance, pvalue > 0.01 hence we conclude that there is no sufficient evidence to deny the claim that the population mean is N$ 5.00