Answer to Question #161233 in Quantitative Methods for Sunny

Question

The mean fat content of a brand of butter is 20.0 mgs. A new process is proposed to lower the fat content without affecting the flavour. To test the new process16 packets of butter are taken at random and the sample mean fat content is found to be 18.5 mgs with a standard deviation of 2 mgs. Test 5 % level of significance is the new process justified.

Answer 1

We have that

$\mu = 20$

$n = 16$

$\bar x=18.5$

$s=2$

$\alpha=0.05$

$H_0:\mu = 20$

$H_a:\mu <20$

The hypothesis test is left-tailed.

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

The critical value for 5% significance level and 15 df is –1.75

(degrees of freedom df = n – 1 = 16 – 1 = 15)

The critical region is t < –1.75

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{18.5- 20}{\frac{2}{\sqrt {16}}}=-1.5$$

Since –1.5 > –1.7 thus t does not fall in the rejection region we fail to reject the null hypothesis.

There is no sufficient evidence to conclude that a new process lowers the fat content without affecting the flavour.