I need some help!!!! Please!!!!!

Ok so I have gotten pretty far in a Stats expample I had to make up, but I have completely forgotten how to go from here!!!! After this step I am supposed to apply it to the t-table, make a visual of a bell shaped curve and determine whether to accept or reject the null hypothesis. Now...let me tell you, my professor is a stickler and wants it his way or no way! EEKKKS!!!!!! So I am reaching out to all you smart people on here to give me some guidance....Here's what I have so far....ooooooh, and P.S. this is due in 9 hours lol....


The United States Medical Board found women living in poverty give birth to an average weight of babies being 2800 grams (the nations average is 3300 grams so this raised question). The U.S.M.B. would like to conduct a test to see if mothers who live in poverty and are given free prenatal care will have babies who weigh significantly higher and closer to the nations average. Twenty-five mothers all living in poverty were used to conduct this sample, the average weight of their babies were 3075 grams and had a standard deviation of 300 grams. Has the program been effective in raising the birth weight of these babies compared to before and after the prenatal care?
1. Null Hypothesis- H0: μ=2800 grams
Alternate Hypothesis- HA: μ ≥ 2800 grams
2. a= .05 (level of significance)
3. Sample average= 3075 grams
Population average= 2800 grams
Standard Deviation= 300 grams
Sample size= 25
3075g-2800g
---------------= 4.58
300 ÷ √25

Now I am to the question-
-----Compare the observed value of the statistic to the critical value obtained for the chosen a(level of significance)
(Now he wants us to draw out a bell shape curve and show the reject region, which I am sure I can handle that if someone can help me get this one under control....I would greatly appreciate it!!!!)

laceylee said:

1. Null Hypothesis- H0: μ=2800 grams
Alternate Hypothesis- HA: μ ≥ 2800 grams
2. a= .05 (level of significance)
3. Sample average= 3075 grams
Population average= 2800 grams
Standard Deviation= 300 grams
Sample size= 25
3075g-2800g
---------------= 4.58
300 ÷ √25

Now I am to the question-
-----Compare the observed value of the statistic to the critical value obtained for the chosen a(level of significance)

Click to expand...

Under the null hypothesis, the mean is 2800g. The only estimator I see to use for the population standard deviation is the 300g that was observed - that is what you have used in the calculation. For a sample of size 25 you correctly divide by 5. The statistic you have calculated is the number of standard deviations above the mean your sample is.

What is your "chosen level of significance"? It looks like you are to choose a, since it was not given. How far above the mean to you require the result to be before you will reject the null hypothesis? a=90%? a=99%? The "critical statistic" is how high above the mean of the normal distribution you go, such that the tail (just the upper tail) contains only (1-a) of the total.

DrPhil said:

Under the null hypothesis, the mean is 2800g. The only estimator I see to use for the population standard deviation is the 300g that was observed - that is what you have used in the calculation. For a sample of size 25 you correctly divide by 5. The statistic you have calculated is the number of standard deviations above the mean your sample is.

What is your "chosen level of significance"? It looks like you are to choose a, since it was not given. How far above the mean to you require the result to be before you will reject the null hypothesis? a=90%? a=99%? The "critical statistic" is how high above the mean of the normal distribution you go, such that the tail (just the upper tail) contains only (1-a) of the total.

Click to expand...

Thank You for responding!!! I really appreciate it!
Yes, a=.05 which means that my level of significance is 95%. Can you just walk me through step by step how I finish this up and how I am supposed to correct my mathematical error that you see. I am a really good student, but I for some reason am having a really hard time in Business Statistics.

laceylee said:

Yes, a=.05 which means that my level of significance is 95%. Can you just walk me through step by step how I finish this up and how I am supposed to correct my mathematical error that you see.

Click to expand...

oh yes - now I see that a=0.05 WAS given. Your "critical value" of the z-statistic is found by looking at the table of the cumulative normal distribution. You want to say, "There is only a 5% chance that purely random statistics would give a result greater than this distance above the mean." This is a single-tailed confidence limit, because H_A is "greater than," not "different from." Look in the table for x such that F(x)=0.95. If your observed statistic z=4.58 is greater than the critical value, you should reject the null hypothesis at the 95% confidence level.

edit: I didn't notice any arithmetic errors

DrPhil said:

oh yes - now I see that a=0.05 WAS given. Your "critical value" of the z-statistic is found by looking at the table of the cumulative normal distribution. You want to say, "There is only a 5% chance that purely random statistics would give a result greater than this distance above the mean." This is a single-tailed confidence limit, because H_A is "greater than," not "different from." Look in the table for x such that F(x)=0.95. If your observed statistic z=4.58 is greater than the critical value, you should reject the null hypothesis at the 95% confidence level.

Click to expand...

0.95=1.645 on the z-table....where I am so lost is how to identify that on a bell shaped curve model, and where each one goes. I guess I am a bit overwhelmed lately and kind of lost to be honest with you. This is my first statistical word problem I have had to create and solve during my tenure at the university. The first half of stats was a walk in the park, now this semester has been a walk in the park.....until now! Goodness I am running crazy....