Answer to Question #183043 in Statistics and Probability for Richard Sasunathan

Question

1. 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

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq20$

$H_1:\mu<20$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha,$ and the critical value for a left-tailed test is $z_c.$

The rejection region for this left-tailed test is $R=\{z:z<z_c\}.$  

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{19.8-20}{0.5/\sqrt{n}}=-0.4\sqrt{40}\approx-2.5298$$

$\alpha=0.05, z_c=-1.6449>-2.5298$

$\alpha=0.01, z_c=-2.3263>-2.5298$

Since it is concluded that the null hypothesis is rejected, then we can use the significance level $\alpha=0.05$ or $\alpha=0.01.$