Question

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. Mean-6206, Median- 4260.5, Standard Deviation- 7419.46802, Min- 444, Max-42,836
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.Mean-3877, Median- 2698, Standard Deviation- 3950.91464, Min- 323, Max-7684
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. Mean-2972, Median- 2086, Standard Deviation- 3183, Min- 29, Max-15238
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. Mean-1527, Median- 1282, Standard Deviation- 1552.29238, Min- 94, Max-7684

e. propose and conduct your own test of hypothesis about the Birth, Death, Marriage and Divorce data that you have been analyzing. Follow these steps (you can make up whatever mean, median, SD, min and max!)

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

PLEASE TYPE!!!!!!! THANK YOU SO MUCH!!!!!!!!!!!!!!!!!

Answer 1

Hi, We are supposed to answer four subparts at a time. Since you have not mentioned which subpart to answer, So I am answering the first four subparts. Please repost the remaining the subparts that you would like to be answered.

a)

Hypothesis:

Null Hypothesis: Ho: u is equal to 5000

Alternate Hypothesis: Ha: u is greater than 5000

Given Information:

Mean = 6206

Standard Deviation = 7419.46

First we need calculate the value of test statistic to determine the p value:

By referring the standard normal table for one tail, the p-value at z = 0.1625 is 0.4354

Since the p-value is more than the level of significance (0.05), the null hypothesis will be accepted.

No, there is not 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.

b)

Hypothesis:

Null Hypothesis: Ho: u is equal to 6000

Alternate Hypothesis: Ha: u is not equal to 6000

Given Information:

Mean = 3877

Standard Deviation = 3950.91

First we need calculate the value of test statistic to determine the p value:

By referring the standard normal table for two tail, the p-value at z = -0.537 is 0.5961

Since the p-value is more than the level of significance (0.1), the null hypothesis will be accepted.

Yes, 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.

c)

Hypothesis:

Null Hypothesis: Ho: u is equal to 2500

Alternate Hypothesis: Ha: u is greater than or  equal to 2500

Given Information:

Mean = 2972

Standard Deviation = 3183

First we need calculate the value of test statistic to determine the p value:

By referring the standard normal table for one tail, the p-value at z = 0.1482 is 0.4410

Since the p-value is more than the level of significance (0.05), the null hypothesis will be accepted.

No, there is no 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.

d)

Hypothesis:

Null Hypothesis: Ho: u is equal to 4000

Alternate Hypothesis: Ha: u is less than or  equal to 4000

Given Information:

Mean = 1527

Standard Deviation = 1552.289

First we need calculate the value of test statistic to determine the p value:

By referring the standard normal table for one tail, the p-value at z = -1.593 is 0.063

Since the p-value is more than the level of significance (0.1), the null hypothesis will be rejected

Yes, 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.