An organization that packages bananas disclosed that the...

Name: an organization that packages bananas disclosed that the weight standard deviation of their banana box packing is 052 kg a new proposal was passed by the management to use biodegradable pack 80951
Uploaded: 2021-12-27T17:56:32-08:00
Duration: 5 min 38 s
Channel: David Nguyen
Description: an organization that packages bananas disclosed that the weight standard deviation of their banana box packing is 052 kg a new proposal was passed by the management to use biodegradable pack 80951

An organization that packages bananas disclosed that the weight standard deviation of their banana box packing is 0.52 kg. A new proposal was passed by the management to use biodegradable packaging. They are afraid that the new change would result in less consistent packaging. A sample of 15 packages was chosen which had a variance of 0.43 kg. Conduct the appropriate test at α = 0.01 level of significance. Use the classical approach and the P-value approach. For this question, don’t test for normality nor sample size. Write the formulas used and mention whether it is a left-tail, right-tail, or two-tailed test.

Transcript

00:01 Hi, i'm david and i'm here to help you answer your question.

00:03 Now let me bring up your question here.

00:06 In this question we will to discuss about the hypothesis testing using the two approaches.

00:11 The first one using the comfort interval.

00:13 The second one using the b value.

00:16 Now, let's try to summarize the question first.

00:20 We have the sample n equal to 150 and the sample mean export equal to the 118.

00:32 18 .86 sample standard deviation equal to the 3 .7.

00:40 And now we want the company want to test whether the mean amount in less than 20 .5 bouts or not.

00:48 So the noon hypothesis, the mean here will be greater than equal to the 20 .5 versus the alternating one mean will be smaller than the 20 .5.

01:00 Now, we're going to do the map approach 1 here with the confidence interval for the mean so the formula will be the x -par plus or minus z n4 over 2 times s over square to n now for the confident interval with the 10 % the z and 4 over 2 it will equal to the z of the 5 % 5 % equal to 0 05 and it will look the table the 5 % 0 .05 here equals to the 1 .645 and therefore we should be able to obtain the confidence interval here.

01:48 Now xp is equal to the 18 .86 plus and minus 1 .645, ask 3 .7 over the square root of the 150.

02:00 Now we compute the quantity on the right.

02:04 We will have the 1 .645 times 3 .7, divided by square root the 150, equal to the 0 .497...

Question

Please give Ace some feedback