Question

Analyst of a company wants to estimate the real amount of palm oil contained in 1 liter containers purchased by the company. The manufacturer has specified that the standard deviation of the amount of oil is 0.04 liter. A random sample of 75 containers is selected and the sample mean amount of oil per one-liter container per one liter is 0.982 liter.

REQUIRED

1. Construct a 95% confidence interval estimate of the μ amount of oil included in a liter container

2. From this analysis, do you think that the analyst has a reason to complain to the manufacturer? Why?

3. Do you think that the μ amount of oil per liter is normally distributed here? Explain.

4. If the analysis was done with a 99% confidence interval estimate, does it change your answer to (b)?

Answer 1

(a)

Suppose, random variable X denotes amount (in liter) of palm oil contained in a 1 liter container.

We have sample values. But we do not know population standard deviation (or variance). So, we have to use test statistic corresponding to one sample t-test.

Corresponding test statistic is given by

Here,

Sample size

Sample mean

Sample standard deviation

Degrees of freedom

We know,

[Using R-code 'qt(1-(1-0.95)/2,74)']

Hence, 95% confidence interval is given by (0.9727968, 0.9912032).

(b)

Calculated 95% confidence interval does not contain manufacturer's claimed value (1 liter). In fact it contains lesser than claimed value. So, the analyst has a reason to complain to the manufacturer.

(c)

Even if probability distribution of oil contained in a container does not follow Normal distribution, we are taking 75 (>30) containers for our estimation purpose. So, we can apply central limit theorem to approximate probability distribution to Normal one.

(d)

We know,

[Using R-code 'qt(1-(1-0.99)/2,74)']

Calculated 99% confidence interval still does not contain manufacturer's claimed value (1 liter). In fact it contains lesser than claimed value. So, it does not change our answer to (b).