Identify the correct HYPOTHESES used in a hypothesis test of the following claim

ID: 2910053 • Letter: I

Question

Identify the correct HYPOTHESES used in a hypothesis test of the following claim and sample data:
Claim: “Golfers who had one lesson with this instructor had an average increase of more than 15 yards in the distance of their drives.”
A random sample of 5 golfers who had attended a free trial lesson with this golf instructor was selected. Before the lesson, each of these golfers hit several drives from a tee, and the average distance for each golfer was recorded (in yards). This process was repeated after the lesson. Assume that the differences between the “before” and “after” distances are approximately normally distributed. Test the claim at the 0.1 significance level.

a. H0: ?d = 15 vs. H1: ?d < 15
b. H0: ?after ? ?before = 15 vs. H1: ?after ? ?before < 15
c. H0: ?d = 15 vs. H1: ?d > 15
d. H0: ?after ? ?before = 15 vs. H1: ?after ? ?before > 15

Identify the P-VALUE in a hypothesis test of the following claim and sample data:
Claim: “The proportion of symptom-free placebo subjects is less than the proportion of symptom-free subjects who received a new supplement.”
A medical researcher wants to determine if a new supplement is effective at reducing the duration of the common cold. Healthy subjects are intentionally infected with a cold-causing virus, and then given either a placebo or the new supplement one day later. Of the 267 subjects who were given a placebo, 33% were symptom-free a week later. And of the 289 subjects who were given the supplement, 56% were symptom-free after a week. Test the claim at the 0.01 significance level.

a. 0.000 000 022 6 (or 2.26×10-8)
b. 0.024
c. 0.012
d. 0.000 000 079 4 (or 7.94×10-8)

Aaron Buddy Chloe David Eric Dist Before Lesson 201.0 195.1 186.4 236.5 250.4 Dist After Lesson 224.7 208.6 200,6 264.3 261.9 Golfer

Explanation / Answer

Solution:

Part 1

Here, we have to use a paired t-test for significant differences in population means.

The null and alternative hypothesis for this test is given as below:

H0: µafter - µbefore = 15 vs. H1: µafter - µbefore > 15

(Correct answer: d)

This is a one tailed test. (Upper tailed or right tailed)

We are given

Level of significance = ? = 0.10

Test statistic = t = t = (Dbar - µd) / [Sd/sqrt(n)]

From given data, calculation table is given as below:

After

Before

(Di - DBar)^2

224.7

201

23.7

30.9136

208.6

195.1

13.5

21.5296

200.6

186.4

14.2

15.5236

264.3

236.5

27.8

93.3156

261.9

250.4

11.5

44.0896

Dbar = 18.14

Sd = 7.1654

n = 5

df = n – 1 = 5 – 1 = 4

t = (Dbar - µd) / [Sd/sqrt(n)]

t = (18.14 – 15) / [7.1654/sqrt(5)]

t = 0.9799

P-value = 0.1913

(by using t-table)

P-value > ? = 0.10

So, we do not reject the null hypothesis

There is no sufficient evidence to conclude that Golfers who had one lesson with this instructor had an average increase of more than 15 yards in the distance of their drives.

Part 2

Here, we have to use z test for population proportions.

H0: p1 = p2 vs. H1: p1 < p2

We are given

? = 0.01

P1 = 0.33

P2 = 0.56

N1 = 267

N2 = 289

Test statistic is given as below:

Z = (P1 – P2) / sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Z = (0.33 – 0.56) / sqrt[(0.33*(1 – 0.33)/267)+(0.56*(1 – 0.56)/289)]

Z = -0.23/ sqrt((0.33*(1 - 0.33)/267)+(0.56*(1 - 0.56)/289))

Z = -5.61028

P-value = 0.0000000226 (correct answer: a)

P-value = 0.00 approximately

P-value < ? = 0.01

So, we reject the null hypothesis

There is sufficient evidence to conclude that the proportion of symptom-free placebo subjects is less than the proportion of symptom-free subjects who received a new supplement.