Question

a. what is the test static?

t=?

b. what is the observed significance level?

p-value=?

c. find the 95% confidence level for (u1-u2).

the confidence interval is (__,__)

Nutritix + Course X PDo Hot X lli MyStatx E P Value X Compu X » C mathxl.com/Student/PlayerHomework.aspx?hom... ☆ I Apps Google A Hanle Econ 2C Summer 2020 section 80 Homework: CH 8.2 Homework Save Score: 0.5 of 1 pt 3 of 3 (3 complete) HW Score: 77.78%, 2.33 of 3.. 8.2.7-T Question Help Independent random samples from normal populations produced the results shown in the table to the right. Complete parts a through d below. Sample 1 1.2 1.4 1.4 1.8 1.5 Sample 2 3.1 3.6 3.1 2.7 a. Calculate the pooled estimate of o? = 0856 (Round to four decimal places as needed.) b. Do the data provide sufficient evidence to indicate that Hy ? Test using a =0.10. What are the null and alternative hypotheses? O A Ho: ---0.10 H:H-4220.10 C. Ho 4-2-0.10 H:-H250.10 B. How-H2 = 0 H:M-H2 <0 OD. Ho: 14-20 H-430 What is the best statistic? (Round to two decimal places as needed) # al va More Enter your answer in the answer box and then click Check Answer parts remaining Clear A Check Answer MacBook P

Answer 1

Ans : Given data are

sample1	sample2
1.2	3.1
1.4	3.6
1.4	3.1
1.8	2.7
1.5

a)

Calculate the pooled estimate of $\large \sigma ^2$

For sample1

sample mean = $\large \bar{x}$ 1 = 7.3/5

$\large \bar{x}$1 = 1.46

sample standard deviation

S= (E(Xi – X)^2 n - 1

S = (E(Xi – 1.46)^2 4

s = 0.2190

for sample 2

sample mean = $\large \bar{x}$ 2 = 12.5/4

$\large \bar{x}$2 = 3.125

sample standard deviation

S= (E(Xi – X)^2 n - 1

S = (E(Xi – 3.125)^2 3

s= 0.3685

the pooled estimate of $\large \sigma ^2$

Sp? (n1 – 1)s12 + (n2 – 1)s22 nl + n2 - 2

_{$\large ^{Sp^2}$ = 0.0856}

a)

Ans : The null and alternative hypothesis is

To Test :

H₀= µ₁ - µ₂ =0 versus   H_a : µ₁ - µ₂ <0

Ans :B

Test Statistic

t = 21 - 22-A (n1 - 1)s12 + (n2-)s22 nl + n2 - 2 1 n1 1 n2 +

1 - 2 - A t = 1 Sp + 1 n2 Vn1

1.46 - 3.125 t = 1 0.2926 * + 5 1 4

t= -8.48

b) p - value

$\large \alpha$ = 0.10

df = n1+n2-2 = 5+4-2 = 7

p- value =  0.000031

critical region : If p - value < $\large \alpha$ then reject H0

since p-value < $\large \alpha$ =0.10 then we reject the null hypothesis at 0.1 level of significance

Conclusion : We conclude that there is sufficient evidence to indicate that the  µ2 > µ₁ at 0.1 level of significance

c)

To find the 95% confidence level for (u1-u2).

100(1-$\large \alpha$)% Confidence interval for two populaton mean difference is

1 1 (1 – 82 +ta, n1 + n2 – 2 * Sp + n1 n2

the 95% confidence level for (u1-u2).

1 1 (1.46 – 3.125 + 10.1,7 * 0.2926* + 5 4

(-1.665 +1.415 + 0.1963)

(-1.9427 , -1.3872)

the confidence interval is (-1.9427 , -1.3872)