Question #348566

You are given the following information of a company that produces cola fizzy drinks. The

company states that the mean caffeine content per bottle of a fizzy drink is 40 mg with a

standard deviation of 7.5 mg. The quality controller is convinced that it is lower. A sample of

30 randomly drawn bottles has a mean caffeine content of 39.2 mg. Can the quality

controller reject the claim? Conduct a hypothesis test at 0.05 level of significance.


1
Expert's answer
2022-06-07T09:26:21-0400

The following null and alternative hypotheses need to be tested:

H0:μ40H_0:\mu\ge40

H1:μ<40H_1:\mu<40

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 α=0.05,\alpha = 0.05, and the critical value for a left-tailed test is zc=1.6449.z_c = -1.6449.

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

The z-statistic is computed as follows:


z=xˉμσ/n=39.2407.5/30=0.5842z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{39.2-40}{7.5/\sqrt{30}}=-0.5842

Since it is observed that z=0.5842>1.6449=zc,z=-0.5842>-1.6449=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=P(z<0.5842)=0.279543,p=P(z<-0.5842)= 0.279543, and since p=0.279543>0.05=α,p= 0.279543>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is less than 40, at the α=0.05\alpha = 0.05 significance level.


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