Can i get help on the problem below, I'm having trouble where the question marks are.

This is a ONE-sample hypothesis test of means using one of the variables from a study. Use ana (or "alpha") = .05. (You may do either a one or two tailed test - use the one that makes the most interesting question to you.)

State the question you are attempting to answer with your hypothesis test in plain English. Hint: you "test the theory" of what is in Ho and "try to prove" what is in H1.

Ho: equal to 40

H1: is not equal to 40

Show the 7 steps for this test in your answer.By the way you do not have to draw the picture of the rejection regions in step 4. If you get step 5 correct, then I will be able to figure out what your picture would have looked like! HINT:If your sample size is greater than 30 you use the z table and if it is less than 31 you use the t table.

Insert 7 steps and SPSS output by step 6 here. Make sure your spss output has the two boxes illustrated.

One-Sample Statistics
	N	Mean	Std. Deviation	Std. Error Mean
Age (in exact years)	35	41.8571	13.63418	2.30460

One-Sample Test
	Test Value = 40
t	df	Significance	Mean Difference	95% Confidence Interval of the Difference
One-Sided p	Two-Sided p	Lower	Upper
Age (in exact years)	.806	34	.213	.426	1.85714	-2.8264	6.5406

Step 1: State the null and alternative hypothesis (H0 and H1).

Ho: equal to 40

H1: is not equal to 40

Step 2: State level of significance or "alpha."

For this example, we'll use =.05

Step 3: Determine the test distribution to use - z or t.

For this example, although the population parameters are unknown, we have a sample size bigger than 30 so we use z.

Step 4 and 5

In this case, we have one tailed test so all 1% goes into the LEFT or NEGATIVE tail. That

translates to z ????? NEED THIS Z =WHAT

Step 5: State the decision rule.

Reject the null if the TR< ????? otherwise FTR. ???

Step 6: SPSS TR says=.806

Step 7: Compare TR value with the decision rule and make a statistical decision.

+2.13falls in the rejection region. Therefore, we reject the null and can conclude that the mean age of 95% confidence.

P-value in plain English p=100% .213% = 2.13% or.??? p< therefore reject the null hypothesis.

In this case P=.213 or 2.13% P= 100% - ???%=.213 p