Question

In a random sample of​ males, it was found that 20 write with their left hands and 222 do not. In a random sample of​ females, it was found that 64 write with their left hands and 439 do not. Use a 0.01 significance level to test the claim that the rate of​ left-handedness among males is less than that among females. Complete parts​ (a) through​ (c) below.

Answer 1

n1 = 20 + 222 = 242

n2 = 64 + 439 = 503

x1 = 20, x2 = 64

= 0.01

= 0.083, = 0.127

= 0.113

The null and alternative hypothsis is

H0: p1=p2

H1: p1<p2

Calculate the test statistic.

test statistics: z = -1.80

Calculate the P-value.

P-Value = P(Z < -1.80)

using normal z table we get

P(Z < -1.80)= 0.0357

P-value = 0.036

decision rule is

Reject Ho if ( P-value ) ( )

here, ( P-value= 0.036) > ( = 0.01)

Hence, we can say,

Do not reject null hypothesis.

Conclusion:

The P-value greater than the significance level of = 0.01, so fail to reject the null hypothesis. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.