Answer to Question #130756 in Statistics and Probability for HALIM

Question

A company sells packets of durian crackers that they claim on average, contains at least 20g of durian crackers. They admit that their claim is wrong (and will refund any money), if a sample of 40 durian crackers have a mean less than 19.8g. If the standard deviation is 0.5g, determine the critical value that specifies the rejection region.

Answer 1

As claimed by the company let us consider the null hypothesis to be,

H₀ : $\mu \ge 19.8$ against the alternate hypothesis H₁ : $\mu < 19.8$

where $\mu$ = population mean weight of the durian crackers.

But since this is a left tailed test the rejection region is (- $\infin$, Z $(1 - \alpha)$]

Let us assume the level of significance = 0.05

So the value of Z $(1 - \alpha)$ from the standard normal distribution table we get,

Z $(1 - 0.05)$ = -1.6449

Hence the rejection region becomes (- $\infin$, -1.6449]