3304.2 8 620.82 s Use a 5% significance level to test the claim that the standard deviation of birth weights of girls is different from the standard deviation of birth weights of boys, which is 470 g. Round all answers to 3 decimal places if possible. Procedure: One variance y* Hypothesis Test v Step 1. Hypotheses Set-Up: Hạ: G v= 470 H: G 470 The test is a two-taled v test. Step 2. The significance level is a = 0.05 Step 3. Compute the value of the test statistic: X 61.087 Step 4. Testing Procedure: Provide the critical value(s) for the Rejection Region. For a one-tailed test, use DNE for the unneeded critical value. • Left CV= Right CV = The p-value is Link to Chi-Square Table: https://www.itl.nist.govidiv898/handbook/eda/section3/eda3674.htm Step 5. Decision Is the test statistic in the rejection region?| Is the p-value less than the significance level? | Conclusion: Select an answer Step 6. Interpretation: At a 5% significance level we Select an answer v have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.
Permutations and Combinations
If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.
Counting Theory
The fundamental counting principle is a rule that is used to count the total number of possible outcomes in a given situation.
can you please do:
Step 4 , Step 5 and Step 6
Step 5: yes or no
Step 6: do or do not
![A sample of birth weights of 36 girls was taken. Below are the results (in g):
3134.1 3503.4 3380.3 3131.9
3638.6
4574.4
3907.2
3913.6
2629.9
3440 3901.8
3047.9
3369.6
2728.1
4921.4 2827.3
2943.6
3044
2614.2
2970.8
2592.9
3314.2
4282.9
3761.8
3489.8
3890.7
3434
3378.8
2405.6
2518.8
3377.8
4080
2097.5
| 2976.1
2923
2803.8
I= 3304.2 8
8 = 620.82 8
Use a 5% significance level to test the claim that the standard deviation of birth weights of girls is different
from the standard deviation of birth weights of boys, which is 470 g.
Round all answers to 3 decimal places if possible.
Procedure: One variance x Hypothesis Test V
Step 1. Hypotheses Set-Up:
Ho: G
v = 470
Hạ: G
470
The test is a two-tailed
v test.
Step 2. The significance level is a =
0.05
Step 3. Compute the value of the test statistic: Xe
v= 61.067
Step 4. Testing Procedure:
Provide the critical value(s) for the Rejection Region. For a one-tailed test, use DNE for the
unneeded critical value.
• Left CV =
• Right CV =
The p-value is
O Type here to search](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F48526f08-4776-4d2d-90ba-eb5cf64fe34e%2F7506a4e2-fdca-4b97-93f3-b10da2901172%2Fz667a2t_processed.png&w=3840&q=75)
![1 = 3304.2 g
8 = 620.82 g
Use a 5% significance level to test the claim that the standard deviation of birth weights of girls is different
from the standard deviation of birth weights of boys, which is 470 g.
Round all answers to 3 decimal places if possible.
Procedure: [One variance x' Hypothesis Test v
Step 1. Hypotheses Set-Up:
Ho: G
v = 470
470
The test is a two-tailed
v test.
Step 2. The significance level is a =
0.05
Step 3. Compute the value of the test statistic: x
v= 61.067
Step 4. Testing Procedure:
Provide the critical value(s) for the Rejection Region. For a one-tailed test, use DNE for the
unneeded critical value.
• Left CV =
• Right CV =
The p-value is
Link to Chi-Square Table: https://www.itl.nist.gov/div898/handbook/eda/section3/eda3674.htm
Step 5. Decision
Is the test statistic in the rejection region? [? v
Is the p-value less than the significance level? ?
Conclusion: Select an answer
Step 6. Interpretation:
At a 5% significance level we Select an answer v have sufficient evidence to reject the null
hypothesis in favor of the alternative hypothesis.
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