A machine is used to fill packets with rice. The contents of each packet weighs X grams - Normal Distribution

Question

A machine is used to fill packets with rice. The contents of each packet weighs X grams. X is normally distributed with mean μ grams and standard deviation 3.71 grams.

The mean weight μ is stated to be 500 g. To check this statement, a sample of 13 packets is selected, and the mean contents x-bar is calculated. The hypotheses for the test are:

H0: μ = 500 g H1: μ≠ 500 g

The critical region is defined by {x ̅<498}∪{(x ) ̅>502}

a) What is the nature of the distribution of X?

My answer: X~N(500, 3.71^2)

b)Show that the significance level for the test is approximately 0.0519

(xbar)~N(500, ((3.71)/√13)^2) P(x >= 502) = 0.0259 0.0259*2 = 0.0518

Doubt is in part (c):

c)Find the probability of a Type II error with this test given that the true value of μ is 498.4 grams.

My answer:

I'm using Casio fx-CG50 (graphic calculator) and:

P(Reject H1 | H1 true):

x-bar ~N(498.4, 3.71/√13)

P(498< x-bar < 502) = 0.651

I am unsure if this is the correct approach for part c.

Answer 1

After a discussion with @Mike,

The answer I've given:

P(498< x-bar < 502) = 0.651

Is correct.

When finding the probability for type 2 error, we must find the probability of x-bar falling in the critical region because when it falls in the critical region we must reject H1.

We take the mean for the x-bar distribution to be the true mean which in this case is: 498.4

and the standard deviation to be which was calculated before using the formula:

Answer 2

Type 2 error would be $P(\text{Not reject } H_0| H_0 \text{ false})$

$=P(\frac{498-498.4}{3.71}<\frac{X-498.4}{3.71}<\frac{502-498.4}{3.71})=P(-0.1078<Z<0.9704)\cong0.83398-0.4562=0.37778$