Question

Biologist in modern days usually use the term “oxygen demand” to explain the amount of oxygen that is needed by aquatic organism to survive life. A group of biological scientist conducted a study in a particular swam area in order to know the average mean of oxygen needed by this aquatic organism in that area for survival. The scientist decided to take a random sample of 45 aquatic organisms in this area and measure the oxygen needed for their survival in mg/L. Conduct a confidence interval for true mean of oxygen needed for survival in this area. The data are shown below in mg/L. 8.4 7.9 9.2 9.1 8.6 9.0 9.5 9.4 8.9 8.9 9.4 8.5 8.7 8.8 9 9.3 9.5 9.5 8.6 8.9 8.4 7.9 9.2 9.5 9.6 9.2 9.1 8.6 8.7 8.5 9.2 8.7 8.8 8.6 9.2 8.8 7.9 8.9 9.1 9.5 8.8 8.9 9.2 9.1 9.4

Answer 1

Answer : 95% confidence interval for true mean = ( 8.8013 , 9.0609 )

Explanation:

To find the confidence interval for true mean, consider the 95% confidence level (i.e alpha = 0.05 )

Now the fromula for confidence interval is

Where is sample mean, n is sample size and s is sample standard deviation.

t is table value from t distribution table with n-1 df and alpha = 0.05

Here we use t distribution because the population standard deviation is unknown.

Summary statistics of given data in excel

Column1
Mean	8.931111
Standard Error	0.064418
Median	8.9
Mode	9.2
Standard Deviation	0.432131
Sample Variance	0.186737
Kurtosis	0.219814
Skewness	-0.65558
Range	1.7
Minimum	7.9
Maximum	9.6
Sum	401.9
Count	45

from this = 8.931111 and s = 0.432131, n = 45

t = 2.0154 ... ( from table with df = 44 and alpha = 0.05)

Also can be obtained in excel =TINV(0.05,44) = 2.0154

= (8.801281 , 9.060938 ) = ( 8.8013 , 9.0609 ) ... rounded to 4 decimals

MINITAB output for reference:

One-Sample T: k

Variable N Mean StDev SE Mean 95% CI
k 45 8.9311 0.4321 0.0644 (8.8013, 9.0609)

Same output is obtained.