Please Send me its solution if some one have..........

Ten oil tens are taken at random from an automatic filling machine the mean weight of the tins is 15.8 kg and the standard deviation is 0.50 kg. Does the sample mean differ siginificantly from the intended weight of 16 kg?

2. Originally Posted by sitara
[snip]
Ten oil tens are taken at random from an automatic filling machine the mean weight of the tins is 15.8 kg and the standard deviation is 0.50 kg. Does the sample mean differ siginificantly from the intended weight of 16 kg?
You can flesh out the argument .....

Since the population variance is not known, the test statistic to use is $t = \frac{\bar{x} - 16}{s/\sqrt{n}}$ which may be compared with Student's t-distribution with (n - 1) degrees of freedom.

$t = \frac{15.8 - 16}{0.50/\sqrt{9}} = -1.2$.
The critical value is -2.262 at the $\alpha = 0.025$ significance level. Therefore ......

Alternatively, a 95% confidence interval for $\mu$ is given by $\bar{x} \pm 2.262 \frac{s}{\sqrt{n}}$ where in this problem $\, \bar{x} = 15.8 \,$, $\, s = 0.50 \,$ and n = 10.
95% confidence interval: $15.8 \pm 0.377$. Therefore .....

The critical value is got using the t-distribution with 10 - 1 = 9 degrees of freedom. The t-distribution is is used because the population standard deviation is not known and the sample size is small (n < 50).