Question

An industrial supplier has shipped a truckload of Teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of these is "in excess" of the 12 ounces printed on each cartridge. To check this claim, a sample of n = 20 cartridges is randomly selected from the shipment and carefully weighed. Summary statistics for the sample are: Sample mean = 12.11 Sample standard dev = 0.30 ounce To determine whether the supplier's claim is true, consider the test Ho: μ = 12 vs. Ha: μ > 12, where μ is the true mean weight of the cartridges. A) Find the critical value for this test. (No significance level provided in the problem) A. 4.67 B. 1.729 C. -1.64 D. 1.725 E. 0 B) Calculate the p-value for this test. A. 0.05 < P < 0.10 B. 0.0505 C. 0.10 < P < 0.20 D. The correct answer is not among the choices. E. 0.1010

          An industrial supplier has shipped a truckload of Teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of these is "in excess" of the 12 ounces printed on each cartridge. To check this claim, a sample of n = 20 cartridges is randomly selected from the shipment and carefully weighed. Summary statistics for the sample are: 
Sample mean = 12.11 
Sample standard dev = 0.30 ounce 
To determine whether the supplier's claim is true, consider the test 
Ho: μ = 12 vs. Ha: μ > 12, where μ is the true mean weight of the cartridges.

A) Find the critical value for this test. (No significance level provided in the problem)
A. 4.67
B. 1.729
C. -1.64
D. 1.725
E. 0

B) Calculate the p-value for this test.
A. 0.05 < P < 0.10
B. 0.0505
C. 0.10 < P < 0.20
D. The correct answer is not among the choices.
E. 0.1010

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Added by Lisa J.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert David Nguyen

07/08/2022

Step 1

This is calculated using the formula: Test statistic = (sample mean - population mean) / (sample standard deviation / sqrt(sample size)) In this case, the test statistic = (12.11 - 12) / (0.30 / sqrt(20)) = 1.47 A) The critical value for this test depends on Show more…

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An industrial supplier has shipped a truckload of Teflon lubricant cartridges to an aerospace customer. The customer has been assured that the mean weight of these is "in excess" of the 12 ounces printed on each cartridge. To check this claim, a sample of n = 20 cartridges is randomly selected from the shipment and carefully weighed. Summary statistics for the sample are: Sample mean = 12.11 Sample standard dev = 0.30 ounce To determine whether the supplier's claim is true, consider the test Ho: μ = 12 vs. Ha: μ > 12, where μ is the true mean weight of the cartridges. A) Find the critical value for this test. (No significance level provided in the problem) A. 4.67 B. 1.729 C. -1.64 D. 1.725 E. 0 B) Calculate the p-value for this test. A. 0.05 < P < 0.10 B. 0.0505 C. 0.10 < P < 0.20 D. The correct answer is not among the choices. E. 0.1010

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Transcript

00:01 Hi, i'm david and i'm here to have you on seeing your question.

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

00:06 In the question here, we are going to discuss about the hypothesis testing.

00:10 Here we will test the null hypothesis with the mean equal to the 12, and the alternative hypothesis mean will be greater than 12.

00:17 And notice that this will be the right -tale test.

00:21 And in this question we're given the sample size and it will equal to the 20.

00:29 And we're given the x -par equal to the 3.

00:31 12 .11 and the sum of standard division s equal to 0 .3 and the question asks me to find a the critical value for the test.

00:43 Here we're given the unfath in this question here it will be equal to when you find the critical value for the test no significant level provided in this problem so i will assume that the unfath equal to the 5 percent which is equal to the 0105.

01:08 And because we are given, this will be the sample standard divisions.

01:14 It means that the sigma is unknown.

01:18 When sigma is unknown, we have to use the t test.

01:22 Therefore, from here we can find the critical value.

01:25 It will equal to the t and then n minus 1 degree freedom.

01:31 And far.

01:32 Here n equals to the 20, minus 1 will be the 19 degree freedom.

01:35 And unfur in this case question equal to the 0 .05.

01:40 And if you want to look up the t table, let me copy the t table...

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