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19 May, 05:08

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 404.0404.0 gram setting. It is believed that the machine is underfilling the bags. A 4848 bag sample had a mean of 400.0400.0 grams. A level of significance of 0.020.02 will be used. Is there sufficient evidence to support the claim that the bags are underfilled

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  1. 19 May, 07:16
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    Bags are being underfilled by the machine

    Step-by-step explanation:

    Solution:-

    - The potato chips manufacturer wants to statistically determine whether the machine is operating correctly based on its weight control.

    - The weight control limit was placed on the machine for 404.0 grams.

    - He believes that the machine is underfilled.

    - So, he takes a sample of size n = 48 bags and weigh them separately and get the following data for his sample.

    Sample Mean (x^) = 400.0 grams

    Variance (σ) = 169 grams

    - He wants to statistically check his doubts on the machine operability at a significance level (α) of 0.02.

    - We will make an hypothesis that the machine works fine. So,

    Null hypothesis: u = 404.0 grams

    - We will test against what he believes to be the case i. e " under-filling of bags "

    Alternate hypothesis: u < 404.0 garms

    - Next we will determine what test is to be applied in this case.

    Population variance (σ^2) is known Sample size n > 30

    - For the above two conditions we can apply the use standard normal distribution (Z-score).

    - We are testing for underfilling of bags. So we will apply one-left tail test.

    - Hence, the critical value for the Alternate hypothesis to hold true would be:

    P (Z < Z-critical) = α

    P (Z < Z-critical) = 0.02

    Z-critical = - 2.06

    - Next we compute the Z-statistics associated with the sample data obtained:

    Z-test = (x^ - u) / √ (σ^2 / n)

    Z-test = (400 - 404) / √ (169 / 48)

    Z-test = - 4 / 1.87638

    Z-test = - 2.13176

    - For the left-one tailed test the Z-critical value provide the limit underneath which all the statistical test values lie in the rejection region.

    Therefore,

    -2.13176 < - 2.06

    Z-test < Z-critical. (Null hypothesis Rejected)

    Conclusion:

    - The statistics value lies in the rejection region; hence, the Null hypothesis is rejected. With can conclude with 98% confidence that the claim made on bags being underfilled by the machine is evident.
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