A bottling machine in a chemical plant supposed to fill average of 250 ml of acids and the data is approximately normally distributed a stan

Question

A bottling machine in a chemical plant supposed to fill average of 250 ml of acids and the data is approximately normally distributed a standard deviation of 20 ml. Plant inspector checked machines annually by taking a sample of 12 acid bottles and computing the average content. If the sample average falls in the interval 240 ml and 260 ml, the machine is considered to be operating satisfactorily and he conclude that y = 250 ml.
(a) Find the probability of committing a typel error.
(b) Find the probability of committing a typell error when = 242 ml.

in progress 0
Savannah 1 month 2021-09-11T10:17:27+00:00 1 Answer 0

Answers ( )

    0
    2021-09-11T10:19:01+00:00

    Answer:

    Check the explanation

    Step-by-step explanation:

    (a) P(Type I error) = P( H_0 is rejected |

    Here we have  : µ = 250 and : µ ≠ 250

    Here it is given that if 240 < X < 260 then the machine is considered operating satisfactory i.e.  is not rejected.

    So rejection region is X <= 240 and X >= 260.

    Also we know that X follows N( µ , (400/12) ) as population follows N( µ, 400) ; So ( X – µ) / (20 / (12)1/2) follows std. normal distribution.

    So, P( is rejected | is true) = P(X <= 240 | µ = 250) + P(X >= 260 | µ = 250)

    = P(( X – µ) / (20 / (12)1/2) <=(240 -250)/ (20 / (12)1/2| µ = 250) +

    P(( X – µ) / (20 / (12)1/2) >=(260 -250)/ (20 / (12)1/2| µ = 250)

    = P(( X – µ) / (20 / (12)1/2) <= -1.7320508076 | µ = 250) + P(( X – µ) / (20 / (12)1/2) >= 1.7320508076 | µ = 250)

    And since ( X – µ) / (20 / (12)1/2) follows std. normal distribution. This can be found from Z-tables.

    so, = 0.0416 + 0.0416 = 0.0832

    (b) P(Type II error) = P( H_0 is not rejected | H_0 is false) = P(240 < X < 260 | µ ≠ 250)

    We have to calculate probability of type II error specifically at µ = 242 so,

    = P(240 < X < 260 | µ =242) = P((240 -242)/ (20 / (12)1/2 <= ( X – µ) / (20 / (12)1/2) <= (260 -242)/ (20 / (12)1/2)| µ = 242)

    = P(-0.3464101615 <= ( X – µ) / (20 / (12)1/2) <= 3.1176914535 | µ = 242)

    And since ( X – µ) / (20 / (12)1/2) follows std. normal distribution. This can be found from Z-tables.

    = 0.6346

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )