Question

5. The greenhouse emissions of 32 cars were measured with a mean of 7.78 tons and a standard deviation of 1.08 tons. (The amounts are in tons per year, expressed a CO2 equivalents.) A government report states that all cars have a mean greenhouse gas emission of 8.0 tons. Test the report's claim. a) State H? and H? for the test. b) Assume ?=0.05. Draw a picture of the distribution, find the critical value(s), and label the rejection regions c) Compute the sample test statistic and locate it on your picture above. d) Based on the above, circle your conclusion: REJECT H? or DO NOT REJECT H? e) State your conclusion in non technical terms. f) Explain in the context of the problem above what would have to happen to be guilty of committing type I error. [Use no symbols and do not use any phrase that would include "null hypothesis" or "alternative hypothesis"!!]

          5. The greenhouse emissions of 32 cars were measured with a mean of 7.78 tons and a standard deviation of 1.08 tons. (The amounts are in tons per year, expressed a CO2 equivalents.) A government report states that all cars have a mean greenhouse gas emission of 8.0 tons. Test the report's claim.

a) State H? and H? for the test.

b) Assume ?=0.05. Draw a picture of the distribution, find the critical value(s), and label the rejection regions

c) Compute the sample test statistic and locate it on your picture above.

d) Based on the above, circle your conclusion: REJECT H? or DO NOT REJECT H?

e) State your conclusion in non technical terms.

f) Explain in the context of the problem above what would have to happen to be guilty of committing type I error. [Use no symbols and do not use any phrase that would include "null hypothesis" or "alternative hypothesis"!!]

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5. The greenhouse emissions of 32 cars were measured with a mean of 7.78 tons and a standard deviation of 1.08 tons. (The amounts are in tons per year, expressed a CO2 equivalents.) A government report states that all cars have a mean greenhouse gas emission of 8.0 tons. Test the report's claim.

a) State H? and H? for the test.

b) Assume ?=0.05. Draw a picture of the distribution, find the critical value(s), and label the rejection regions

c) Compute the sample test statistic and locate it on your picture above.

d) Based on the above, circle your conclusion: REJECT H? or DO NOT REJECT H?

e) State your conclusion in non technical terms.

f) Explain in the context of the problem above what would have to happen to be guilty of committing type I error. [Use no symbols and do not use any phrase that would include "null hypothesis" or "alternative hypothesis"!!]

Added by Todd G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Madhur L

05/09/2023

Step 1

0 tons. The alternative hypothesis is that the mean emissions for some cars is different from 8.0 tons. The sample test statistic is t=1.08, which is located in the rejection region (shown by the red circle) for the alternative hypothesis. Show more…

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The greenhouse emissions of 32 cars were measured with a mean of 7.78 tons and a standard deviation of 1.08 tons. (The amounts are in tons per year; expressed as CO2 equivalents). A government report states that all cars have a mean greenhouse gas emission of 8.0 tons. Test the report's claim. State Ho and Ha for the test. Assume α = 0.05. Draw a picture of the distribution, find the critical value(s), and label the rejection regions. Compute the sample test statistic and locate it on your picture above. Based on the above, circle your conclusion: REJECT Ho or DO NOT REJECT Ho State your conclusion in non-technical terms. Explain in the context of the problem above what would have to happen to be guilty of committing type I error.

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Transcript

00:01 Hello students, we have given the sample size n is equal to 32 then sample mean x bar is equal to 7 .78 and sample standard deviation that is s is equal to 1 .08 and we have to test the reports claim that all the cars have the mean greenhouse gas emission of 8 tons that is we have to test the null hypothesis h0 as mu is equal to 8 versus the alternative hypothesis ha as mu is not equal to 8 and this we have to test at significance level alpha is equal to 0 .05.

00:41 Now here we have the sample size n is equal to 32.

00:44 So therefore the degrees of freedom is equal to n -1 which is equal to 31 and therefore the critical value that is tc is equal to t alpha by 2, degrees of freedom which is equal to t 0 .025, 31 and by using the t table this value is 2 .04.

01:07 Now by using the normal curve we can show the rejection region.

01:13 So here the point is critical value which is minus 2 .04 and on the right side it is positive 2 .04.

01:25 So the area to the left hand side of minus 2 .04 and the region to the right hand side of the positive 2 .04.

01:38 This is the area of the rejection.

01:41 So this is rejection region, rejection region.

01:49 Then similarly this shaded part is also the rejection region, rejection region and now in the c part we have to compute the sample test statistic and locate it in the this normal curve.

02:09 So here we have given the value of s that is the sample standard deviation and population standard deviation is unknown.

02:18 So therefore we will be using the t distribution test and the test statistic for that is given by t which is equal to x bar minus mu divided by s upon square root of n.

02:31 So therefore now this is equal to 7 .78 minus 8 divided by 1 .08 divided by square root of 32 and that is equal to minus 1 .15...

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