Question

Consider the following hypotheses:

H₀: μ = 4,500
H_A: μ ≠ 4,500

The population is normally distributed with a population standard deviation of 700. Compute the value of the test statistic and the resulting p-value for each of the following sample results. For each sample, determine if you can "reject/do not reject" the null hypothesis at the 10% significance level. (You may find it useful to reference the appropriate table: z table or t table) (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round "test statistic" values to 2 decimal places and "p-value" to 4 decimal places.)

		Test statistic	p-value
a.	x−x− = 4,630; n = 135			(Click to select)  Do not reject H0  Reject H0
b.	x−x− = 4,630; n = 265			(Click to select)  Reject H0  Do not reject H0
c.	x−x− = 4,210; n = 50			(Click to select)  Reject H0  Do not reject H0
d.	x−x− = 4,290; n = 50			(Click to select)  Do not reject H0  Reject H0

Answer 1

Solution :

= 4500

= 700

a) = 4630

n = 135

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 4500

H_a : 4500

Test statistic = z

= ( - ) / / n

= (4630-4500) / 700 / 135

= 2.16

p(Z > 2.16) = 1-P (Z <2.16 ) =0.0309

P-value = 0.0309

= 0.10

p=0.0309<0.10,

Reject the null hypothesis .

b)

= 4630

n = 265

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 4500

H_a : 4500

Test statistic = z

= ( - ) / / n

= (4630-4500) / 700 / 265

= 3.02

p(Z > 3.02) = 1-P (Z <3.02 ) =0.0025

P-value = 0.0025

= 0.10

p=0.0025<0.10

Reject the null hypothesis .

c)

= 4210

n = 50

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 4500

H_a : 4500

Test statistic = z

= ( - ) / / n

= (4210-4500) / 700 / 50

= -2.93

P (Z <-2.93 ) =0.0034

P-value = 0.0034

= 0.10

p=0.0034<0.10

Reject the null hypothesis .

d)

= 4290

n = 50

This is the two tailed test .

The null and alternative hypothesis is

H₀ :   = 4500

H_a : 4500

Test statistic = z

= ( - ) / / n

= (4290-4500) / 700 / 50

= -2.12

P (Z <-2.12) =0.0339

P-value = 0.00339

= 0.10

p = 0.0339<0.10

Reject the null hypothesis .