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Sagot :
There is no sufficient evidence to support the claim that there is a difference in the longevity of the two brands of batteries , as the null hypothesis failed to be rejected.
So the mean and standard deviation for the sample of battery 1 is :
M(1)= 106+ 111+ 109+ 105/4 =107.75
standard deviation (1)=r√(x(i)-M/(n-1)=root(22.75/3)=2.754
Similarly, for the second battery, the mean and standard deviation is
M(2)= 125 +103+ 121+ 118/4=116.75
S(2)=√(276.75/3)=√(92.25)=9.605
Here the hypothesis test for the difference between populations means, here the claim is that the longevity of the two brands of batteries differs.So the alternative and null hypotheses are:
H(0)= μ1-μ2=0
H(1)=μ1-μ2 ≠0
Here the significance level is 0.05(α), The difference between sample means is M(d)=107.75- 116.75=-9 .Now the estimated standard error of the difference between the means is :
s(Md)=√(2.754²+9.605²/4)=4.996, Now the test static will be, t=-9-0/4.996=-1.8014 and the degree of freedom is 4+4-2=6 .Here the test is two-tailed with degrees of freedom of the test static is -1.8014, so the P-value for the test is : 2.P(t< -1.8014)=0.122
Here the P-value (0.122) is greater than the significance level so the effect is insignificant, and the null hypothesis failed to be rejected.
To know more about the hypothesis refer to the link brainly.com/question/28920252
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the operations manager of a manufacturer of television remote controls wants to determine which batteries last the longest in his product. he took a random sample of his remote controls and tested two brands of batteries. the number of minutes of continuous use before the batteries failed for each brand is given in battery.xlsx. if the significance level is 10%, is there statistical evidence of a difference in longevity between the two batteries?
The sample data for each battery is:
Battery 1: 106 111 109 105
Battery 2: 125 103 121 118
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