Question #339698

2. A light-bulb manufacturer regularly advertises that his bulbs last 900 hours


with a standard deviation of 75 hours. A random sample is chosen before each


campaign to make sure that the claim is correct. If one such sample of 20 bulbs


show a mean of 925 hours, can the advertising claim be considered an


underestimate at the 0.05 level of significance?


1
Expert's answer
2022-05-13T04:50:11-0400

The following null and alternative hypotheses need to be tested:

H0:μ900H_0:\mu\le 900

Ha:μ>900H_a:\mu>900

This corresponds to a right-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 right-tailed test is zc=1.6449.z_c = 1.6449.

The rejection region for thisright-tailed test is R={z:z>1.6449}.R = \{z: z >1.6449\}.

The z-statistic is computed as follows:


z=xˉμσ/n=92590075/201.4907z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{925-900}{75/\sqrt{20}}\approx1.4907

Since it is observed that z=1.4907<1.6449=zc,z = 1.4907 <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>1.4907)=0.06802,p=P(Z>1.4907)=0.06802, and since p=0.06802>0.05=α,p=0.06802>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 greater than 900, at the α=0.05\alpha = 0.05 significance level.


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