Question

An official of a large national union claims that the fraction of women in the union is not significantly different from 58%. To test this claim, a random sample of 400 individuals in the union is taken. Of those 400,215 are women. Use the p-value to conduct the hypothesis test and use a =0.05 level of significance. p-value = 0.0446 <0.05; Fail to reject H. - O a. p-value = 0.0446 <0,05; Reject H. Ob. p-value = 0.4714 > 0.05; Fail to reject Ho Ос. p-value = 0.4714 > 0.05; Reject He Od

Answer 1

1) The null and alternative hypotheses are:

Ho:p=0.58 and Hp0.58 O b.

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2)

Proportion of women , p = 215/400 = 0.5375

Sample size, n = 400

P0 = 0.58

Standard error, se = sqrt(P0*(1-P0)/n)

= sqrt(0.58*(1-0.58)/400)

= 0.0247

z = (P - P0)/se = (0.5375-0.58)/0.0247 = -1.7

p-value = NORMSDIST(-1.7) = 0.0446

P-value is less than 0.05, therefore, null hypothesis is rejected.

p-value = 0.0446 <0.05; Reject H. Ob.

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