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.

ur mom

step-by-step explanation: