A sample of birth weights of 35 girls was taken. Below are the results (in g): 2991.9 3538.6 3780.4 2622.2 3601.7 3833.8 3497.5 3665.8 3702.2 3771.9 3308.9 3428.5 2857.1 3422.9 3218.2 3052.4 3320.1 3380.3 2989.5 3642 3383.3 4131.4 3194.3 3298.4 4536.2 3290.4 3479.9 4385.3 3828.1 3165.7 2976.7 2877.2 4162.4 3239.9 2957 (Note: The average and the standard deviation of the data are respectively 3443.8 g and 435.92 g.) Use a 5% significance level to test the claim that the standard deviation of birthweights of girls is less than the standard deviation of birthweights of boys, which is 500 g. Procedure: Select an answer One mean T Hypothesis Test One proportion Z Hypothesis Test One variance χ² Hypothesis Test One mean Z Hypothesis Test Assumptions: (select everything that applies) Sample size is greater than 30 Population standard deviation is unknown The number of positive and negative responses are both greater than 10 Population standard deviation is known Simple random sample Normal population Step 1. Hypotheses Set-Up: H0:H0: Select an answer p μ σ² = , where ? σ p μ is the Select an answer population proportion population standard deviation population mean and the units are ? mg g lbs kg Ha:Ha: Select an answer σ² p μ ? > < ≠ , and the test is Select an answer Two-Tailed Left-Tailed Right-Tailed Step 2. The significance level α=α= % Step 3. Compute the value of the test statistic: Select an answer χ₀² t₀ z₀ = (Round the answer to 3 decimal places) Step 4. Testing Procedure: (Round the answers to 3 decimal places) CVA PVA Provide the critical value(s) for the Rejection Region: Compute the P-value of the test statistic: left CV is and right CV is P-value is Step 5. Decision: CVA PVA Is the test statistic in the rejection region? Is the P-value less than the significance level? ? no yes ? yes no Conclusion: Select an answer Reject the null hypothesis in favor of the alternative. Do not reject the null hypothesis in favor of the alternative. Step 6. Interpretation: At 5% significance level we Select an answer DO DO NOT have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis

A sample of birth weights of 35 girls was taken. Below are the results (in g): 2991.9 3538.6 3780.4 2622.2 3601.7 3833.8 3497.5 3665.8 3702.2 3771.9 3308.9 3428.5 2857.1 3422.9 3218.2 3052.4 3320.1 3380.3 2989.5 3642 3383.3 4131.4 3194.3 3298.4 4536.2 3290.4 3479.9 4385.3 3828.1 3165.7 2976.7 2877.2 4162.4 3239.9 2957 (Note: The average and the standard deviation of the data are respectively 3443.8 g and 435.92 g.) Use a 5% significance level to test the claim that the standard deviation of birthweights of girls is less than the standard deviation of birthweights of boys, which is 500 g. Procedure: Select an answer One mean T Hypothesis Test One proportion Z Hypothesis Test One variance χ² Hypothesis Test One mean Z Hypothesis Test Assumptions: (select everything that applies) Sample size is greater than 30 Population standard deviation is unknown The number of positive and negative responses are both greater than 10 Population standard deviation is known Simple random sample Normal population Step 1. Hypotheses Set-Up: H0:H0: Select an answer p μ σ² = , where ? σ p μ is the Select an answer population proportion population standard deviation population mean and the units are ? mg g lbs kg Ha:Ha: Select an answer σ² p μ ? > < ≠ , and the test is Select an answer Two-Tailed Left-Tailed Right-Tailed Step 2. The significance level α=α= % Step 3. Compute the value of the test statistic: Select an answer χ₀² t₀ z₀ = (Round the answer to 3 decimal places) Step 4. Testing Procedure: (Round the answers to 3 decimal places) CVA PVA Provide the critical value(s) for the Rejection Region: Compute the P-value of the test statistic: left CV is and right CV is P-value is Step 5. Decision: CVA PVA Is the test statistic in the rejection region? Is the P-value less than the significance level? ? no yes ? yes no Conclusion: Select an answer Reject the null hypothesis in favor of the alternative. Do not reject the null hypothesis in favor of the alternative. Step 6. Interpretation: At 5% significance level we Select an answer DO DO NOT have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A sample of birth weights of 35 girls was taken. Below are the results (in g):

2991.9	3538.6	3780.4	2622.2	3601.7
3833.8	3497.5	3665.8	3702.2	3771.9
3308.9	3428.5	2857.1	3422.9	3218.2
3052.4	3320.1	3380.3	2989.5	3642
3383.3	4131.4	3194.3	3298.4	4536.2
3290.4	3479.9	4385.3	3828.1	3165.7
2976.7	2877.2	4162.4	3239.9	2957

(Note: The average and the standard deviation of the data are respectively 3443.8 g and 435.92 g.)

Use a 5% significance level to test the claim that the standard deviation of birthweights of girls is less than the standard deviation of birthweights of boys, which is 500 g.

Procedure: Select an answer One mean T Hypothesis Test One proportion Z Hypothesis Test One variance χ² Hypothesis Test One mean Z Hypothesis Test

Assumptions: (select everything that applies)

Sample size is greater than 30
Population standard deviation is unknown
The number of positive and negative responses are both greater than 10
Population standard deviation is known
Simple random sample
Normal population

Step 1. Hypotheses Set-Up:

H0:H0: Select an answer p μ σ² =	, where ? σ p μ is the Select an answer population proportion population standard deviation population mean and the units are ? mg g lbs kg
Ha:Ha: Select an answer σ² p μ ? > < ≠	, and the test is Select an answer Two-Tailed Left-Tailed Right-Tailed

Step 2. The significance level α=α= %

Step 3. Compute the value of the test statistic: Select an answer χ₀² t₀ z₀ = (Round the answer to 3 decimal places)

Step 4. Testing Procedure: (Round the answers to 3 decimal places)

CVA	PVA
Provide the critical value(s) for the Rejection Region:	Compute the P-value of the test statistic:
left CV is and right CV is	P-value is

Step 5. Decision:

CVA	PVA
Is the test statistic in the rejection region?	Is the P-value less than the significance level?
? no yes	? yes no

Conclusion: Select an answer Reject the null hypothesis in favor of the alternative. Do not reject the null hypothesis in favor of the alternative.

Step 6. Interpretation:

At 5% significance level we Select an answer DO DO NOT have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

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