Question #349955

Of all of the individuals who develop a certain rash, suppose the mean recovery time for individuals who do not use any form of treatment is 30 days with standard deviation equal to 8. A pharmaceutical company manufacturing a certain cream wishes to determine whether the cream shortens, extends, or has no effect on the recovery time. The company chooses a random sample of 100 individuals who have used the cream, and determines that the mean recovery time for these individuals was 28.5 days. Does the cream have any effect?

1
Expert's answer
2022-06-14T00:01:13-0400

The following null and alternative hypotheses need to be tested:

H0:μ30H_0:\mu\ge30

H1:μ<30H_1:\mu<30

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=28.5308/100=1.875z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{28.5-30}{8/\sqrt{100}}=-1.875

Since it is observed that z=1.875<1.6449=zc,z=-1.875<-1.6449=z_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is p=P(z<1.875)=0.030396,p=P(z<-1.875)=0.030396, and since p=0.030396<0.05=α,p=0.030396<0.05=\alpha, it is concluded that the null hypothesis is rejected.

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

is less than 30, at the α=0.05\alpha = 0.05 significance level.


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