Question #346290

A company produces mosquito repellants that have a period of effectiveness that is approximately normally distributed with a mean of 300 hours and a standard deviation of 30 hours. A sample of 40 repellants was taken and the average period of effectiveness is computed as 290 hours. Is this sufficient evidence to conclude that the company’s claim is true at α=0.01?

1
Expert's answer
2022-05-31T11:26:23-0400

The following null and alternative hypotheses need to be tested:

H0:μ=300H_0:\mu=300

H1:μ300H_1:\mu\not=300

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=29030030/40=2.1082z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{290-300}{30/\sqrt{40}}=-2.1082


6. Since it is observed that z=2.1082<2.5758=zc,|z|=2.1082<2.5758=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed is p=2P(Z<2.1082)=0.035014,p=2P(Z<-2.1082)=0.035014, and since p=0.035014>0.01=α,p=0.035014>0.01=\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 different than 300, at the α=0.01\alpha = 0.01 significance level.


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