Answer to Question #169724 in Statistics and Probability for satya

Question

A small scale restaurant owner wants to find an estimate of the average daily profit. For this, he randomly selected 25 days from the past year’s entry log and collected the following observations:

a. Sample mean of the daily profits for those 25 selected days = Rs. 33892

b. Sample standard deviation of the daily profits for those 25 selected days = Rs. 435

Assuming that daily profit values follow normal distribution, answer the following questions:

a) What is the target population? [1]

b) What is the population parameter that needs to be estimated? [1]

c) Find a 95% confidence interval for the population mean (average daily profit). [4]

d) How do you interpret this confidence interval obtained in 2.c)? [2]

e) Find a 95% confidence interval for the population mean (average daily profit) if the sample size is 100 instead of 25. [2]

Answer 1

We have that

$n = 25$

$\bar x=33892$

$s=435$

a) the target population is all days the small scale restaurant is working

b) the population parameter needed to be estimated is the population mean

c) 95% confidence interval for the population mean is calculated by the formula:

$$\bar x\pm t_{\frac{\alpha}{2},df}\frac{s}{\sqrt n}$$

where df = n – 1 = 25 – 1 = 24

$33892\pm 2.064 \cdot \frac{435}{\sqrt 25}=33892\pm180$

95% confidence interval for the population mean is (34072; 33712)

d) We are 95% confident to conclude that the population mean is located in the interval (34072; 33712).

e) 95% confidence interval for the population mean if the sample size is 100:

n = 100

$$\bar x\pm t_{\frac{\alpha}{2},df}\frac{s}{\sqrt n}$$

where df = n – 1 = 100 – 1 = 99

$33892\pm 1.984 \cdot \frac{435}{\sqrt 25}=33892\pm172$

95% confidence interval for the population mean is (33720; 34064)