Question

uppose a national news agency reported that more than 88% of all households purchased at least one item online last year. We wish to test this statement and randomly sample 200 households. When asked if they made any online purchases last year 180 reported “YES”. At the 0.05 level of significance, does the sample data support the statement made by the national news agency? 1. State the null and alternative hypotheses of interest. Ho ___________________________ HA ___________________________

2. What is the critical value of the test statistic at the α = 0.05 level of significance?

3. What is the calculated value of the test statistic computed from the sample data?

4. State your conclusion. Does the sample data support the statement made by the news agency?

Answer 1

Answer)

1)

Null hypothesis Ho : P = 0.88

Alternate hypothesis Ha : P > 0.88

2)

N = 200

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 = 176

N*(1-p) = 24

Both the conditions are met so we can use standard normal z table to estimate the critical value

From z table, P(z>1.645) = 0.05

So, Z = 1.645

Rejection region is reject Ho if test statistics is greater than 1.645

3)

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

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

Observed P = 180/200

Claimed = 0.88

N = 200

Z = 0.87

4)

As 0.87 is not greater than 1.645

We fail.to reject the null hypothesis Ho

We do not have sufficient evidence to support the claim.that P > 0.88