Question

A sample of 60 observations is selected from one population with a population standard deviation of 0.68. The sample mean is 268. A sample of 49 observations is selected from a second population with a population standard deviation of 0.69. The sample mean is 2.62. Conduct the following test of hypothesis using the 0.1 significance level: HO H-120 th: H1-H2 > 0 a. Is this a one-tailed or a two-tailed test? This is a one -tailed test. b. State the decision rule. (Round the final answer to 2 decimal places.) The decision rule is to reject Ho if zis greater than c. Compute the value of the test statistic (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.) The test statistic is z d. What is your decision regarding Ho? Ho is not rejected e. What is the p-value? (Round the final answer to 4 decimal places.) The p-value is

Answer 1

Answer:

n1= 60,
C
= 2.68,  $1591303245674_blob.png$= 0.68,

n2= 49,  
T'
= 2.62,  $1591303245652_blob.png$ = 0.69

$1591303307518_blob.png$ = 0.1

Ho:
ul
- $1591303245654_blob.png$$\leq$ 0

H1:
ul
- $1591303245654_blob.png$ > 0

a)

This is one tailed test.

b)

find z critical value for right tailed test with $1591303307518_blob.png$ = 0.1

using normal z table we get

critical value = 1.28

The decision rule is reject Ho if z is greater than 1.28

c)

$z =\frac{\bar x1 -\bar x2}{\sqrt{\frac{\sigma 1^{2}}{n1} + \frac{\sigma 2^{2}}{n2}}}$

$z =\frac{2.68 - 2.62 }{\sqrt{\frac{0.68^{2}}{60} + \frac{0.69^{2}}{49}}}$

z = 0.454558

The test statistics: Z= 0.45

d)

since, ( z = 0.45) < 1.28

Null hypothesis is not rejected.

e)

P-Value = 1 - P(Z < 0.45)

where, P(Z < 0.45) = 0.6753

P-Value = 1 -  0.6753

P-Value = 0.3247