In a random sample of males it was found that 17 write with

In a random sample of males, it was found that 17 write with their left hands and 209 do not. In a random sample of females, it was found that 63 write with their left hands and 433 do not. Use a 0.05 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. Identify the test statistic. (Round to two decimal places as needed.) Identify the P-value. P-value = [] (Round to three decimal places as needed.) What is the conclusion based on the hypothesis test? the significance level of a = 0.05, so The P-value is the null hypothesis. There evidence to support the claim that the rate of left-handedness among males is less than that among females.. Test the claim by constructing an appropriate confidence interval. The 90% confidence interval is Q]

Solution

Proportion of left handers in males p_m= 17/(17+209) = 0.075

n₁ = 17 + 209 = 226

Proportion of left handers in females p_f= 63/(63+433) = 0.128

n₂ = 63 + 433 = 496

Null hypothesis: P_m = P_f
Alternative hypothesis: P_m < P_f

Pooled sample proportion p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)

                       = (0.075 * 226 + 0.128 * 496) / (226 + 496)

                       = 0.11

Std Error = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] } = (0.11 * (1 - 0.11) [ 1/226 + 1/496 ] = 0.025

Z = (P_m - P_f)/SE = (0.075 - 0.128)/0.025 = -2.06

p value = 0.020

p value is less than the significance level so reject the null hypothesis. there is enough/sufficient evidence to support the claim

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alpha = 0.10

Z value for alpha = 0.10 is + 1.65

Margin of error = std error * critical value = 0.025 * + 1.65 = + 0.041

Confidence interval = ((P_m - P_f) - 0.041 ; (P_m - P_f) + 0.041)

= (-0.093, -0.011)

Because the confidence limits is less than 0 , it appears that the two rates of left handedness are significant. There is enough/sufficient evidence to support the claim