The manufacturer of television tubes knows from the past experience that the average life of tube is 2000 hrs with a s.d. of 200 hrs. A sample of 100 tubes 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 2000 hrs.
In this question, we determine whether the population mean"(\\mu)" is equal to 2000hrs as claimed. To do so, we shall test the following hypotheses,
"H_0:\\mu=2000"
"Against"
"H_1:\\mu\\not=2000"
We are given the following information,
"n=100"
"\\bar{x}=1950"
"\\sigma=200" and "\\alpha=0.01"
To perform this test, we shall use the standard Normal distribution since our sample size is large. We proceed as follows.
The test statistic is given as,
"Z^*=(\\bar{x}-\\mu)\/(\\sigma\/\\sqrt{n})"
"Z^*=(1950-2000)\/(200\/\\sqrt{100})=-50\/20=-2.5"
The test statistic "Z^*" is compare with the table value at "\\alpha=0.01" significance level. This table value is given as,
"Z_{\\alpha\/2}=Z_{0.01\/2}=Z_{0.005}=2.575" and the null hypothesis is rejected if "|Z^*|\\gt Z_{0.005}"
Since "|Z^*|=|-2.5|=2.5\\lt Z_{0.005}=2.575", we fail to reject the null hypothesis and we conclude that there is sufficient evidence to show that this sample came from a normal population of mean 2000 hours at 1% level of significance.
Comments
Leave a comment