Question #342403

A chemical company alleged that the average weight of the bag of chemical is 30kgs. with a standard deviation of 1.1 kgs. A sample of 26 bags was taken and revealed a mean weight of 28.9 kgs. Shall we accept the allegation of the chemical company? Use .01 level of significance



Expert's answer

The following null and alternative hypotheses need to be tested:

H0:μ=30H_0:\mu=30

H1:μ30H_1:\mu\not=30

This corresponds to a two-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 α=0.01,\alpha = 0.01, and the critical value for a two-tailed test is zc=2.5758.z_c = 2.5758.

The rejection region for this two-tailed test is R={z:z>2.5758}.R = \{z:|z|>2.5758\}.

The z-statistic is computed as follows:


z=xˉμσ/n=28.9301.1/265.09902z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{28.9-30}{1.1/\sqrt{26}}\approx-5.09902

Since it is observed that z=5.09902>2.5758=zc,|z|=5.09902>2.5758=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=2P(z<5.09902)=0,p=2P(z<-5.09902)=0, and since p=0<0.01=α,p= 0<0.01=\alpha, it is concluded that the null hypothesis is rejected.

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

is different than 30, at the α=0.01\alpha = 0.01 significance level.



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