Answer to Question #332472 in Statistics and Probability for Abbi

Question

A company has developed a new battery. The engineering department of the company claims that each battery lasts for 200 minutes. In order to test this claim, the company selects a random sample of 100 new batteries so that this sample has a mean of 190 minutes. Given that the population standard deviation is 30 minutes, test the engineering department’s claim that the new batteries run with an average of 200 minutes. Use 1% level of significance.

Which is the correct null hypothesis that can be derived in the situation?

Which is the correct alternative hypothesis that can be derived in the situation? 

What test will be used based from the given values in the situation? 

What is the decision based from the critical value and the computed test statistic?

Answer 1

$H_0: \mu=200$

$H_1: \mu\not =200$

Since this is a two tailed test to obtain the critical value we will divide the level of significance by two .We obtain 0.005 .The corresponding z value to the area 0.005 and 0.995 is -2.58 and 2.58.Thus the critical values are -2.58 and 2.58

$Z=\frac{x-\mu}{\sigma}=\frac{190-200}{30}=-3.333$

Z calculated (-3.333) is less than Z critical value (-2.58) thus we reject the null hypothesis at 1% level of significance.

There is sufficient evidence to conclude that new battery run on a time that is significantly different from the claimed 200 minutes.