Question

A sample of 66 observations is selected from one population with a population standard deviation of 0.68. The sample mean is 2.67. A sample of 45 observations is selected from a second population with a population standard deviation of 0.68. The sample mean is 2.59. Conduct the following test of hypothesis using the 0.1 significance level:

*H*_{0}: μ_{1} – μ_{2}≤ 0

*H*_{1}: μ_{1} – μ_{2} > 0

**a.** Is this a one-tailed or a two-tailed
test?

This is a (Click to select) one two -tailed test.

**b.** State the decision rule. **(Round the
final answer to 2 decimal places.)**

The decision rule is to reject *H*_{0} if
*z* is (Click to select) less
than greater than equal
to .

**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
*H*_{0}?

*H*_{0} is (Click to
select) not
rejected rejected .

**e.** What is the *p*-value? **(Round
the final answer to 4 decimal places.)**

The *p*-value is .

Answer #1

a)

This is a one tailed test

b)

Rejection Region

This is right tailed test, for α = 0.1

Critical value of z is 1.28.

Hence reject H0 if z > 1.28

The decision rule is to reject H0 if z is greater than equal to 1.28

c)

Pooled Variance

sp = sqrt(s1^2/n1 + s2^2/n2)

sp = sqrt(0.4624/66 + 0.4624/45)

sp = 0.1315

Test statistic,

z = (x1bar - x2bar)/sp

z = (2.67 - 2.59)/0.1315

z = 0.61

d)

H0 is not rejected

e)

P-value Approach

P-value = 0.2709

p value is calculated By using excel:

= 1 - NORM.DIST(0.61,0,1,TRUE)

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