The operations manager of a manufacturer of television remote controls wants to determine which batteries last the longest in his product. H

Question

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. Here are the number of minutes of continuous use before the batteries failed for each brand. If significance level is 10%, Is there statistical evidence of a difference in longevity between the two batteries?

Battery 1 106 111 109 105

Battery 2 125 103 121 118

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Julia 2 days 2021-09-12T19:55:21+00:00 1 Answer 0

Answers ( )

  1. Emma
    0
    2021-09-12T19:56:47+00:00

    Answer:

    t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

    The degrees of freedom are given by:

    df=n_{1}+n_{2}-2=4+4-2=6

    The p value for this case would be given by:

    p_v =2*P(t_{(6)}<-1.804)=0.121

    Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

    Step-by-step explanation:

    Information given

    Battery 1 106 111 109 105

    Battery 2 125 103 121 118

    We can calculate the mean and the deviation with the following formulas”

    \bar X =\frac{\sum_{i=1}^n X_i}{n}

    s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

    \bar X_{1}=107.75 represent the mean for the Battery 1

    \bar X_{2}=116.75 represent the mean for the Bettery 2

    s_{1}=2.75 represent the sample standard deviation for the Battery 1

    s_{2}=9.604 represent the sample standard deviation for the battery 2

    n_{1}=4 sample size selected for the Battery 1

    n_{2}=4 sample size selected for the Battery 2

    \alpha=0.1 represent the significance level

    t would represent the statistic  

    p_v represent the p value

    System of hypothesis

    We want to check if the difference in longevity between the two batteries, the system of hypothesis would be:

    Null hypothesis:\mu_{1} = \mu_{2}

    Alternative hypothesis:\mu_{1} \neq \mu_{2}

    The statistic is given by:

    t=\frac{\bar X_{s1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

    The statistic is given by:

    t=\frac{107.75-116.75}{\sqrt{\frac{2.75^2}{4}+\frac{9.604^2}{4}}}}=-1.802  

    The degrees of freedom are given by:

    df=n_{1}+n_{2}-2=4+4-2=6

    The p value for this case would be given by:

    p_v =2*P(t_{(6)}<-1.804)=0.121

    Since the p value is higher than the significance level we have enough evidence to FAIl to reject the null hypothesis and we can conclude that the true mean is not significantly different between the two types of battery

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