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.
4)Given that the level of significance is 0.01 or 1%, what is/are the critical values?
5)Using the appropriate formula, what is the computed test statistic? (Input your answer using 3 decimal places, example: 1.234 if positive or -1.234 if the answer is negative) *
6)What is the decision based from the critical value and the computed test statistic
7)What is the conclusion?
4) 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
5) z="\\frac{x-\\mu}{\\sigma}"
x=190
"\\mu"=200
"\\sigma=30"
z="\\frac{190-200}{30}"
z=-3.333
6) Z calculated (-3.333) is less than Z critical value (-2.58) thus we reject the null hypothesis at 1% level of significance.
7) There is sufficient evidence to conclude that new battery run on a time that is significantly different from the claimed 200 minutes.
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