Question

The following is a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Test Upper H Subscript 0 Baseline : p Subscript 1 Baseline equals p Subscript 2 vs Upper H Subscript a Baseline : p Subscript 1 Baseline less-than p Subscript 2 when the samples have n Subscript 1 Baseline equals 130 with p Overscript ^ EndScripts Subscript 1 Baseline equals 0.20, and n Subscript 2 Baseline equals 100 with p Overscript ^ EndScripts Subscript 2 Baseline equals 0.22. The standard error of p Overscript ^ EndScripts Subscript 1 Baseline minus p Overscript ^ EndScripts Subscript 2 from the randomization distribution is 0.05. Find the value of the standardized z-test statistic.

Answer 1

Answer:

Here we have to test the hypothesis that,

H0 : mu = 85 Vs H1 : mu > 85

where mu is population mean.

Assume alpha = level of significance = 0.05

Given that,

n = 19

Xbar = 82.1

s = 3.9

Now we have to find test statistic.

The test statistic is,

Z = (Xbar - mu) / (s/sqrt(n))

= (82.1 - 85) / (3.9/sqrt(19))

= -3.24

Now we have to find P-value for taking decision.

P-value we can find using EXCEL.

syntax :

=1 - NORMSDIST(z)

where z is test statistic.

P-value = 0.999

P-value > alpha

Accept H0 at 0.05 level of significance.

COnclusion : There is sufficient evidence to say that the population mean is 85.

Exercise 054 :

Here we have to test the hypothesis that,

H0 : p1 = p2 Vs H1 : p1 < p2

where p1 and p2 are two population proportions.

Assume alpha = 0.05

Given that,

n1 = 160

p1^ = 0.18

n2 = 100

p2^ = 0.25

Now we can find x by using equation :

x1 = n1*p1^ = 160*0.18 = 28.8

x2 = n2*p2^ = 100*0.25 = 25

Now we have to find best estimate p^.

p^ = x1+x2 / n1+n2

p^ = (28.8+25) / (160+100) = 53.8/260 = 0.207

q^ = 1 - p^ = 1 - 0.207 = 0.793

The test statistic is,

Z = (p1^ - p2^) / sd

where sd = sqrt[(p^*q^) / n1 + (p^*q^)/n2]

sd = sqrt [(0.207*0.793) / 160 + (0.207*0.793) / 100] = 0.052

Z = (0.18 - 0.25) / 0.052 = -1.36