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a) We will check this by finding the confidence interval.
n = 100, x = 44
p̅ = x/n = 0.44
95% Confidence interval :
At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960
Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.44 - 1.96 *√(0.44*0.56/100) = 0.343
Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.44 + 1.96 *√(0.44*0.56/100) = 0.537
0.343 < p < 0.537
As the value 0.33 is not in the interval so we will reject 0.33
As the value 0.34 is in the interval so we will not reject 0.34
The value 0.53 is included in the so we will not reject 0.53
As the value 0.54 is not in the interval so we will reject 0.54
Do not reject the null hypothesis for the values of p0 between 0.34 and 0.53, inclusively.
Do Not Solve. please delete A blind taste test is conducted to determine which of two...
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