Question

Use the following notations for your population values: m₁ are the old golf balls and m₂ are the new ones. Pretend this is a management report although in the form of an answer sheet.

1. List the hypotheses for the tests:
   1. Numeric hypotheses
      
      
   2. Narrative hypotheses (should match above but in English)

2. Before you draw the sample

a. Is it a dependent (paired) or independent sample?

b. What is the z or t value (rejection rule) you will test for at .05?

c. Why did you use z or t?

d. Based on the above what is your rejection rule for the null hypothesis? Describe in a few sentences and explain 2b to your “test statistic”.

Current	New
264.00	277.00
261.00	269.00
267.00	263.00
272.00	266.00
258.00	262.00
283.00	251.00
258.00	262.00
266.00	289.00
259.00	286.00
270.00	264.00
263.00	274.00
264.00	266.00
284.00	262.00
263.00	271.00
260.00	260.00
283.00	281.00
255.00	250.00
272.00	263.00
266.00	278.00
268.00	264.00
270.00	272.00
287.00	259.00
289.00	264.00
280.00	280.00
272.00	274.00
275.00	281.00
265.00	276.00
260.00	269.00
278.00	268.00
275.00	262.00
281.00	283.00
274.00	250.00
273.00	253.00
263.00	260.00
275.00	270.00
267.00	263.00
279.00	261.00
274.00	255.00
276.00	263.00
262.00	279.00

Answer 1

Assuming we want to see if means are different

this is paired t-test

H, :μα = 0

H:Hd 70

Using Excel

data -> data analysis -> t-Test: Paired Two Sample for Means

t-Test: Paired Two Sample for Means
	Current	New
Mean	270.275	267.5
Variance	76.61474359	97.94871795
Observations	40	40
Pearson Correlation	-0.082729738
Hypothesized Mean Difference	0
df	39
t Stat	1.2770
P(T<=t) one-tail	0.1046
t Critical one-tail	1.6849
P(T<=t) two-tail	0.2092
t Critical two-tail	2.0227

t = 1.277

p-value for two-tailed = 0.2092

since p-value > alpha

hence we fail to reject the null hypothesis

we conclude that there is not difference in means