Answer to Question #161249 in Statistics and Probability for Sunny

Question

A random sample of 400 tins of vegetable oil, labelled ""15 kgs net weight"' gave a mean weight of 14.25 kgs with a standard deviation of 1.50 kg. Do we reject the hypothesis of net weight of 15 kgs per tin on the basis of this sample, at 5% level of significance?

Answer 1

We have that

$\mu = 15$

$n = 400$

$\bar x=14.25$

$s=1.50$

$\alpha=0.05$

$H_0:\mu = 15$

$H_a:\mu \ne15$

The hypothesis test is two-tailed.

Since the population standard deviation is unknown we use the t-test.

The critical value for 5% significance level and 399 df is $\pm1.9659$

(degrees of freedom df = n – 1 = 400 – 1 = 399)

The critical region contains all values smaller than –1.9659 and larger than 1.9659

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{14.25- 15}{\frac{1.5}{\sqrt {400}}}=-10$$

Since –10 < –1.9659 thus t falls in the rejection region we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to reject the hypothesis. We are 95% confident to conclude that the mean weight is not 15kgs per tin on the basis of this sample.