Question #274860

The manufacturer of television tube know from the past experience that the average life of tube is 2000 hrs with a s.d. of 200 hrs a simple of 100 tubes has an average life of 1950 hrs test at the 0.01 level of significance to see if this sample come from a normal population of mean 2000 hrs.


Expert's answer

In this question, we determine whether the population mean(μ)(\mu) is equal to 2000hrs as claimed. To do so, we shall test the following hypotheses,

H0:μ=2000H_0:\mu=2000

AgainstAgainst

H1:μ2000H_1:\mu\not=2000

We are given the following information,

n=100n=100

xˉ=1950\bar{x}=1950

σ=200\sigma=200 and α=0.01\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=(xˉμ)/(σ/n)Z^*=(\bar{x}-\mu)/(\sigma/\sqrt{n})

Z=(19502000)/(200/100)=50/20=2.5Z^*=(1950-2000)/(200/\sqrt{100})=-50/20=-2.5

The test statistic ZZ^* is compare with the table value at α=0.01\alpha=0.01 significance level. This table value is given as,

Zα/2=Z0.01/2=Z0.005=2.575Z_{\alpha/2}=Z_{0.01/2}=Z_{0.005}=2.575 and the null hypothesis is rejected if Z>Z0.005|Z^*|\gt Z_{0.005}

Since Z=2.5=2.5<Z0.005=2.575|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.


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