Answer to Question #258572 in Statistics and Probability for deev

Question #258572

the manufacture of television TUBes knows from the past experience thatthe average life of tubes is 2000 hrs with a s.d. of 200hrs . a sample of 100 tube has an average life of 1950 hrs. test at the 0.01 level of significance to see if this sample came from a normal population of mean 2000hrs. step by step explanation


1
Expert's answer
2021-10-31T18:33:26-0400

The following null and alternative hypotheses need to be tested:

"H_0: \\mu=2000"

"H_1:\\mu\\not=2000"

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 "\\alpha = 0.01," and the critical value for a two-tailed test is "z_c=2.5758."

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

The z-statistic is computed as follows:


"z=\\dfrac{\\bar{X}-\\mu}{\\sigma\/\\sqrt{n}}=\\dfrac{1950-2000}{200\/\\sqrt{100}}=-2.5"

Since it is observed that "|z| = 2.5 \\le 2.5758=z_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is "p=2P(Z<-2.5)\\approx0.012419," and since "p=0.012419>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 "\\mu" is different than "2000," at the "\\alpha = 0.01" significance level.

Therefore, there is enough evidence to claim that this sample came from a normal population of mean 2000 hrs, at the "\\alpha = 0.01" significance level.



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