Answer to Question #208689 in Statistics and Probability for amira

Question

According to a report, 30 percent of Malaysian graduates are unemployed. An academician is curious to determine if the same pattern also exists in Kota Kinabalu. He randomly selects 320 non science graduates and 425 science graduates for the analysis.

Field of study Gender Unemployed Employe

Non- science male 54 141

female 53 72

Science male 87 165

female 43 130

a. Test if the proportion of unemployed non-science male graduates is 5 percent less than

the proportion of science male graduates at 10 percent significance level.

i. Specify the competing hypotheses of the test.

ii. Calculate the value of the relevant test statistic.

iii. Compute the p-value.

iv. At 5% significance level, does the proportion of unemployed non-science male

graduates is 5 percent less than the proportion of science male graduates?

Answer 1

Unemployed non-science male graduates = 54

Total non-science male graduates n₁= 54+141 = 195

Proportion of non-science unemployed male graduates:

$\hat{p}_1= \frac{54}{195}=0.2769$

Total science male graduates n₂ = 87 + 165 = 252

Proportion of science unemployed male graduates:

$\hat{p}_2= \frac{87}{252}=0.3452$

i. Z test for two sample proportions

$H_0: p_1-p_2 ≥ -0.05 \\ H_1: p_1-p_2< -0.05$

ii.

Test statistic

$z= \frac{\hat{p}_1 - \hat{p}_2 - (p_1-p_2) }{\sqrt{\bar{p}(1-\bar{p})( \frac{1}{n_1} + \frac{1}{n_2} )}}$

Pooled proportion:

$\bar{p}= \frac{54+87}{195+252}=0.3154 \\ Z= \frac{0.2769-0.3452-(-0.05)}{(0.3154)(1-0.3154)(\frac{1}{195} + \frac{1}{252})} = -0.4129$

iii. Using P-value calculator

Z=-0.4129

For left-tailed test P=0.3398

iv.

α=5%=0.05

0.3398>0.05

p>α

H₀ is accepted.

There is not enough evidence to support the claim that the proportion of unemployed non-science male graduates is 5 percent less than the proportion of science male graduates at 5% significance level.