Answer to Question #308294 in Statistics and Probability for ffi

Question

A vendor claims that the variance for measurement of tiles that his factory produce were 13

square feet. A sample of nine tiles was measured in square feet and the results of the tiles

produced were recorded as below:

204.5 206.3 202.4

207.8 203.1 206.2

203.8 206.6 205.8

(Assuming the sample comes from a normal population)

i) Calculate the point estimate of the population mean.

(2 marks)

ii) Compute the variance and standard deviation.

(3 marks)

iii) Determine the estimate of 95% confidence interval for the population mean of the

tiles.

Answer 1

Given that 204.5,206.3,202.4,207.8,203.1,206.2,203.8,206.6,205.8

i) The point estimate of the population mean = ( (204.5)+(206.3)+(202.4)+(207.8)+(203.1)+(206.2)+(203.8)+(206.6)+(205.8) ) / 9

which yields (1846.5)/ 9 = 205.17 which is the required point estimate of the population mean.

ii) The variance

we define the variance as below

= ( (204.5 -205.17)² +(206.3 -205.17)² + (202.4 - 205.17)² +. . . .+(206.6 - 205.17)² +(205.8-205.17)² ) / (9-1)

solving the above yields

(25.98)/(8) = 3.2475 which is the required variance

The standard deviation = (variance)^1/2 = 1.8021

iii)since the mean is 205.17 , the standard deviation is 1.8021 and the sample size n =9, we define the confidence interval at 0.05 level of significance as below

we define the degree of freedom as ( 9 - 1) =8, we then obtain the critical t value at 0.05 level of significance with 8 degrees of freedom which yields = 2.306

Based on the above information, we obtain the 95% confidence interval as below

lower limit = 205.17 - ( (2.306*1.8021) / (9^1/2) ) = 203.785

upper limit = 205.17 + ( (2.306*1.8021) / (9^1/2) ) = 206.555

Thus we may express the confidence interval as below

(203.785, 206.555)