Question

A random sample of ?=1000 registered voters and found that 520 would vote for the Republican candidate in a state senate race. Let ? represent the proportion of registered voters who would vote for the Republican candidate.
Consider testing

?0:?=.50
??:?>.50

(a) The test statistic is ?z =

(b) Regardless of what you acutally computed, suppose your answer to part (a) was z = 1.28. Using this z, p-value =

Answer 1

Answer)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P > 0.5

N = 1000

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 500

N*(1-p) = 500

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

After substitution

Z = 1.26

B)

From z table, P(z>1.28) = 0.1003

P-value = 0.1003