Question

In cleveland, a sample of 73 mail carriers showed that 10 had received an animal bite during one week. In Philadelphia, in a sample of 80 mail carriers, 16 had recieved animal bites. Is there a significant difference in the proportion? Use a 0.05 significance level. Find the p-value.

Answer 1

Solution :

Given that,

n1 = 73

x1 = 10

Point estimate = sample proportion = 1 = x1 / n1 = 0.137

n2 = 80

x2 = 16

Point estimate = sample proportion = 2 = x2 / n2 = 0.2

The value of the pooled proportion is computed as,

= ( x1 + x2 ) / ( n1 + n2 )

= (10 + 16 ) / (73 + 80 )

= 0.170

1 - = 0.83

Level of significance = = 0.05

This a two-tailed test.

The null and alternative hypothesis is,

Ho: p1 = p2

Ha: p1 p2

Test statistics

z = (1 - 2 ) / *(1-) ( 1/n1 + 1/n2 )

= (0.137 -0.2 ) / (0.170 - 0.83 ) ( 1/73 + 1/80)

= -1.037

P-value = 2*P(Z<z)

=2* P(Z < -1.037)

= 2 * 0.1499

= 0.2998

The p-value is p = 0.2998, and since p = 0.2998 > 0.05, it is concluded that the null hypothesis is fail to rejected.

there a significant difference in the proportion? Use a 0.05 significance level. Find the p-value.

There is no significant difference in the proportion at a 0.05 significance level.