Answer to Question #256353 in Statistics and Probability for anis

Question

QUESTION 1 (15 MARKS)

In a butter-packing factory, the quantity of butter packed in a day using a certain

machine is normally distributed. In a particular day, 12 packets of butter were taken at

random from this production line, and their masses, measured in grams, were:

9.50 9.50 11.20 10.60 9.90 11.10

10.90 9.80 10.10 10.20 10.90 11.00

By using these measurements:

a. Point out the estimator that is used to estimate the mean parameter.

[ 3 marks ]

b. Explore the 95% confidence interval for the mean mass produced by this

machine. [12 marks ]

[Total : 15 marks]

Answer 1

a.

$$\bar{x}=\dfrac{1}{12}(9.50+ 9.50+ 11.20+ 10.60+ 9.90+ 11.10$$

$$+10.90 +9.80 +10.10 +10.20 +10.90+ 11.00)$$

$$=\dfrac{124.7}{12}\approx10.39$$

A point estimate of the mean of a population is determined by calculating the mean of a sample drawn from the population.

Point estimate of population mean

$$\hat{\mu}=\mu_{\bar{x}}=10.39$$

b.

$$s^2=\dfrac{1}{12-1}((9.50-10.39)^2+(9.50-10.39)^2$$

$$+(11.20-10.39)^2+(10.60-10.39)^2$$

$$+(9.90-10.39)^2+(11.10-10.39)^2$$

$$+(10.90-10.39)^2+(9.80-10.39)^2$$

$$+(10.10-10.39)^2+(10.20-10.39)^2$$

$$+(10.90-10.39)^2+(11.00-10.39)^2$$

$$\approx0.399015$$

$$s=\sqrt{s^2}\approx0.631676$$

The critical value for $\alpha = 0.05$ and $df = n-1 = 11$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.200985.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(10.39-2.200985\times\dfrac{0.631676}{\sqrt{12}},$$

$$10.39+2.200985\times\dfrac{0.631676}{\sqrt{12}})$$

$$=(9.989, 10.791)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $9.989 < \mu < 10.791 ,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(9.989, 10.791).$