Answer to Question #183046 in Statistics and Probability for Richard Sasunathan

Question

2. A researcher wishes to test the claim of a particular cereal manufacturer that the mean weight of cereal in the boxes is less than 300g. A sample of 50 boxes yields a sample mean weight of 296g. Assume that the population standard deviation is 5g.

a) Can we conclude that the claim is true? Test at α = 0.05.

b) Obtain a 95% confidence interval for μ.

Answer 1

a) The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq300$

$H_1:\mu<300$

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 $\alpha=0.05,$ and the critical value is $z_c=-1.6449.$

The rejection region for this left-tailed test is $R=\{z:z<-1.6449\}.$  

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{296-300}{5/\sqrt{50}}=-5.6569$$

Since it is observed that $z=-5.6569<-1.6449=z_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 300, at the 0.05 significance level.

Using the P-value approach:

The p-value for left-tailed test, $z=-5.6569$ is $p<0.00001,$ and since $p<0.00001<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 300, at the 0.05 significance level.

b) The critical value for $\alpha=0.05$ is $z_c=z_{1-\alpha/2}=1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-z_c\times\dfrac{\sigma}{\sqrt{n}},\bar{x}+z_c\times\dfrac{\sigma}{\sqrt{n}})$$

$$=(2.96-1.96\times\dfrac{5}{\sqrt{50}},296+1.96\times\dfrac{5}{\sqrt{50}})$$

$$=(294.614, 297.386)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $294.614<\mu<297.386,$ which indicates that we are 95% confident that the true population mean $\mu$  is contained by the interval $(294.614, 297.386).$