Question

Claim: Most adults would erase all of their personal information online if they could. A software firm survey of 561 randomly selected adults showed that 56% of them would erase all of their personal information online if they could. Find the value of the test statistic.

Answer #1

answer)

Null hypothesis Ho : P = 0.5

Alternate hypothesis Ha : P > 0.5 (most means more than 50%)

N = 561

P = 0.5

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 = 280.5

N*(1-p) = 280.5

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

Observed P = 0.56

Claimed P = 0.5

N = 561

After substitution

Test statistics z = 2.84225262775 = 2.84

From z table, P(z>2.84) = 0.0023

P-Value = 0.0023

As the P-Value is extremely small we reject the null hypothesis

and we have enough evidence to support the claim that p > 0.5

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