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.

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.