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.

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.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.