Question

Pick a topic to investigate. Be certain to cite your source(s). You may research a topic about a population mean or a population proportion. Specify whether you are choosing the mean or proportion in your report. Explain why you find this topic interesting. (15 pts.)
Collect or find a source to compare data on the same topic. If you are using a source, it should be cited. Create a confidence interval from the sample. For a sample proportion, use one-prop z interval. For a sample mean, select the T-interval. Be certain to test the conditions. Select your level of confidence: it should be 90%, 95% or 98%. Determine the margin of error for the interval. What is your sample size? How can you decrease your margin of error? (25 pts.)
Test a hypothesis using the value that you obtained in part two about either the true population proportion or true mean. What are your null and alternative hypotheses? Do you have enough evidence to reject the null hypothesis when . (15 pts.)
Explain what the P-value means in the context of the situation. ( 5 points.)
Calculate the test statistic. Can you conclude whether to reject or fail to reject the null hypothesis by using your test statistic? Explain. Show calculations. (10 points.)
When you created your confidence interval in part 3, did that contradict your findings for the hypothesis test? Explain. (10 points)
Write a conclusion regarding your findings. (10 points.)

Answer 1

Summary Table for Live Births
Mean	6206.33
Median	4260.50
Standard Deviation	7419.47
Minimum	444
Maximum	42836

Determine if there is sufficient evidence to conclude the average amount of births is over 8000 in the United States and territories at the 0.05 level of significance.

Clearly state a null and alternative hypothesis.

The null and alternative hypothesis are:

Null hypothesis, Ho: µ = 8000.

The average amount of births is 8000.

Alternative hypothesis, H1: µ> 8000.

The average amount of births is more than 8000.

Give the value of the test statistic.

We are given with:

Sample mean, x = 6206.33

Standard Deviation, s = 7419.47

Significance level, α = 0.05

Sample size, n = 52

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

Test statistic, t = (x - µ)/s/ Ön

                        = (6206.33 – 8000)/7419.47/Ö52

                        = (-1793.67)/7419.47/Ö52

                        = - 1.743

Report the P-Value.

P-value with df = 51 and test statistic = -1.743 at α = 0.05 is, 0.0437.

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

Since p-value is less than the significance level, we can reject the null hypothesis. 0.0437<0.05.

Explain what your conclusion means in context of the data.

Therefore, we have sufficient evidence to say that the average amount of births is more than 8000.