Question

Summary Table for _Live births__for all 52 states___
Mean	302.12
Median	90.5
Standard Deviation	140,98
Minimum	556
Maximum	56,889

Summary Table for _Live births___Deaths__
Mean	306.36
Median	201.5
Standard Deviation	788, 64
Minimum	472
Maximum	39,12

Summary Table for _Live births_____Marriages__
Mean	418.97
Median	155
Standard Deviation	136,74
Minimum	405
Maximum	25, 490

Summary Table for _Live births___Divorces____
Mean	272.72
Median	160
Standard Deviation	8.23
Minimum	222
Maximum	18, 187

Hypothesis Testing

1. Determine if there is sufficient evidence to conclude the average amount of births is over 5000 in the United States and territories at the 0.05 level of significance.

2. Determine if there is sufficient evidence to conclude the average amount of deaths is equal to 6000 in the United States and territories at the 0.10 level of significance.

3. Determine if there is sufficient evidence to conclude the average amount of marriages is greater or equal to 2500 in the United States and territories at the .05 level of significance.

4. Determine if there is sufficient evidence to conclude the average amount of divorces is less than or equal to 4000 in the United States and territories at the 0.10 level of significance.

For each of the tests above, in your report, be sure to—

1. Clearly state a null and alternative hypothesis
2. Give the value of the test statistic
3. Report the P-Value
4. Clearly state your conclusion (Reject the Null or Fail to Reject the Null)
5. Explain what your conclusion means in context of the data.

Lastly, propose and conduct your own test of hypothesis about the Birth, Death, Marriage and Divorce data that you have been analyzing. Make sure to follow the five steps above.

Answer 1

(a)

Clearly state a null and alternative hypothesis.

The null and alternative hypothesis are:

The null hypothesis, Ho: µ = 5000.

The average amount of births is 5000.

The alternative hypothesis, H₁: µ > 5000.

The average amount of births is more than 5000.

Give the value of the test statistic.

We are given with:

Sample mean, x = 302.12

Standard Deviation, s = 140,98

Significance level, α = 0.05

Sample size, n = 52

Degree of freedom, df = n - 1 = 52 – 1 = 51

Test statistic, t = (x - µ)/ s/√n

                        = (302.12 – 5000)/140,98/√52

                        = (-1793.67)/ 7419.47/√52

                        = - 2.403

Report the P-Value.

P-value with df = 51 and test statistic = -1.743 at α = 0.05 is, 0.0100.

Clearly state your conclusion (Reject the Null or Fail to Reject the Null)

Since the p-value is less than the significance level, we can reject the null hypothesis. 0.0100<0.05.

Explain what your conclusion means in the context of the data.

Therefore, we have sufficient evidence to say that the average amount of births is more than 5000.

As per the Chegg answering guide, we have the option to answer only the first question in case of multiple questions. If you want to get the answers for the rest of the parts, please post the question in a new post.

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