Question

QUESTION 1

A report by the Gallup Poll stated that on average a woman visits her physician 5.8 times a year. A

researcher randomly selects 20 women and obtained these data

3 | 2 | 1

3 | 7 | 2

9 | 4 | 6

6 | 8 | 0

5 | 6 | 4

2 | 1 | 3

4 | 1

At α=0.05 can it be concluded that the average is still 5.8 visits per year? Use P-value method.

a) State the hypothesis and identify the claim.

b) State the value of d.f.

c) Compute the test value.

d) Find the P-value.

e) Make the decision

f) Summarize the results

Answer 1

$n = 20 \\ \bar{X} = \frac{3+2+1...+3+4+1}{20}=3.85 \\ s = \sqrt{ \frac{1}{20-1}((3-3.85)^2+(2-3.85)^2+...+(4-3.85)^2+(1-3.85)^2 )}=2.519$

a)

$H_0: \mu = 5.8 \\ H_1: \mu ≠5.8$

b)

d.f. = 20-1=19

c)

$t = \frac{\bar{X} - \mu}{s/ \sqrt{n}} = \frac{3.58-5.8}{2.519/ \sqrt{20}}=-3.46$

d) Since the test is two-tailed, the P-value is the probability that the z-score is smaller than z=−3.46 or larger than z=3.46.

Determine corresponding probability using the normal probability table.

The p-value of a two-tailed test is twice the p-value of the one-tailed test.

$P = 2P(Z< -3.46) \\ = 2 \times 0.00027 \\ = 0.00054$

P-value<0.01

e) P-value=0.01 < α=0.05 reject the null hypothesis

f) There is enough evidence to support the claim that the mean is not 5.8.