Question

HT Proportion: A random sample of 220 individuals who voted in the last election showed that 135 voted for the given amendment. Let be the true proportion of population of voters that voted for the given amendment. Test the claim at the 1% level of significance, that this is two-thirds majority. Determine the p-value to four decimal places.

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.67

Ha : p 0.67

n =220

x =135

= x / n = 135 / 220 =0.61

P0 = 0.67

1 - P0 = 1- 0.67 =0.33

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.61 -0.67 / [(0.67*0.33) / 220]

= -1.78

Test statistic = z = -1.78

P(z < -1.78) = 0.0375

P-value = 2 * 0.0375 =0.0750

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