A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 409 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 401 grams with a standard deviation of 26. A level of significance of 0.1 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

The decision rule is to Reject H0 if Z ≤ -1.282

Step-by-step explanation:

We are given;

Population mean; μ = 409 g

Sample mean; x¯ = 401 g

Sample size; n = 21

Standard deviation; s = 26

Let's define the hypotheses;

Null hypothesis; H0: μ = 409 g

Alternative hypothesis; Ha : μ ≠ 409 g

Formula for test statistic is;

z = (x¯ - μ)/(s/√n)

z = (401 - 409)/(26/√21)

z = -1.410

z-value is negative and thus this is a lower tail test.

At significance level of 0.1, the critical value is -1.282.

Thus, the decision rule is;

Reject H0 if Z ≤ -1.282

(4,5

look up a corrdinate plane and do what the thingy says and boom

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg