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4 April, 17:19

A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 442 gram setting. It is believed that the machine is underfilling the bags. A 44 bag sample had a mean of 438 grams. Assume the population variance is known to be 576. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

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  1. 4 April, 17:30
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    p value is 0.1343

    Step-by-step explanation:

    Null: u> = 442

    Alternative: u < 442

    Using the formula for z score:

    (x - u) / sd/√n

    Where x is 438, u = 442 sd can be determined from the variance = √variance = √576 = 24 and n = 44

    z score = 438-442 / (24/√44)

    z score = - 4 / (24/6.6332)

    z = - 4/3.6182

    z = -1.1055

    Now let's find the p value at 0.1 significance level using a z score of - 1.1055, using a p value calculator, p value is 0.1343 which greatest than 0.1 meaning the day is not sufficient enough to conclude that the machine is underfilling the bags.
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