The manager of a Lay\'s Potato Chip packaging plant is concerned that bags are b

Question

The manager of a Lay's Potato Chip packaging plant is concerned that bags are being overfilled. The mean weight of chips put into the bags on the production line is supposed to be 28.3 grams. The manager selects a random sample of 6 bags from the production line; he carefully weighs the contents of each bag; the weights are as follows (in grams):

29.3, 28.2, 29.1, 28.7, 28.9, 28.5.

Assume that the distribution of the weight of chips put into bags can be approximated with a normal model.

Perform the appropriate hypothesis test at significance level a = .05 and answer the following questions. Calculate the sample mean y and sample standard deviation s to 2 decimal places.

Question 1. What is the value of the test statistic t ?


Question 2. Choose the correct statement concerning the P-value for this test.

P-value > .05 P-value < -1.96 P-value = .05 .025 < P-value < .05 .01 < P-value < .025 P-value < 0.0001 P-value > 1.645

Question 3. What is an appropriate conclusion for this hypothesis test?

a) Reject H0: mu = 28.3; there is sufficient evidence to conclude that the mean fill of the bags is greater than 28.3 grams.
b) Do not reject H0: mu = 28.3; there is insufficient evidence to conclude that the mean fill of the bags is significantly greater than 28.3

Question 4. What type of error might you be making?
4
a) Type I error
b) Type II error

Explanation / Answer

To find the t statistic: t=(mean-h0)/standard error the mean =28.78333 found by adding the 6 values together and dividing by 6 h0= 28.3 (what the bags are supposed to weigh, given in the question) SE= standard dev/square root(n) n=6 (the number of sample bags) standard dev = 0.402078 (found by calculator) so SE= 0.402078/sqroot(6) = 0.164148 plug it all in to find the t statistic = (28.78333 - 28.3)/0.164148 = 2.94 Next find the tails of the p-value by looking at a t-distribution table. On the left you go down to 5 degrees of freedom because degrees of freedom = n-1. Scan the numbers in the 5 df row to find the two numbers that the t statistic lies between. The numbers are .02 and .01 so your answer for p- value is: .01

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