Answer to Question #153097 in Statistics and Probability for rehan tahir

Question

The manager of a bottling plant is anxious to control the mean net weight of fruit bottled. Over a long period, the variance has been 45.2 gm. A new machine is introduced and the net weights (in grams) in randomly selected bottles (all of the same nominal weight) are 987, 966, 955, 977, 981, 967, 975, 980, 953, 972. Would you report to the manager that the new machine has a mean net weight of fruit equal to 975 grams? Also calculate 95% confidence interval for the variance.

Answer 1

$\overline{X}=(987+966+955+977+981+967+975+980+953+972)/10=971.3$

Stating the hypotheses:

H₀: $\mu=975$

H₁: $\mu\ne975$

Computing the test value:

$z=\frac{\overline{X}-\mu}{\sigma/\sqrt{n}}=\frac{971.3-975}{45.2/\sqrt{10}}=-0.26$

Let's find critical value for $\alpha$=0.05 (two-tailed test):

critical values are 1.96 and -1.96

-1.96<-0.26<1.96

So, we can't reject null hypothesis.

We can report to the manager that the new machine has a mean net weight of fruit equal to 975 grams.

Let's calculate 95% confidence interval for the variance:

s² = $\frac{n(\sum{X^2})-(\sum{X})^2}{n(n-1)}=\frac{10*9435347-9713^2}{90}=123.3$

$\frac{(n-1)*s^2}{\chi^2_{right}}<\sigma^2<\frac{(n-1)*s^2}{\chi^2_{left}}$

For $\alpha$=0.05 and d.f.=9:

$\chi^2_{right}$= 16.919

$\chi^2_{left}$= 3.325

$\frac{9*123.3}{16.919}<\sigma^2<\frac{9*123.3}{3.325}$

$65.6<\sigma^2<333.7$