Question

The Mean Corporation manufactures a line of unassembled furniture. Based on historical evidence, the customer service manager knows that the proportion of customers that have not had any problems with assembling the furniture at home is 0.60. Improvements have been made to the instructions provided with the furniture, and the customer service manager believes that this will increase the proportion of customers that can assemble the furniture without problems (p) to above 0.60. A hypothesis test is conducted in order to find out. The null and alternative hypotheses are: H0: p = 0.60 Ha: p > 0.60 To run the test, the manager randomly selects a sample of 101 people and asks them to assemble a piece of furniture produced by the Mean Corporation. The proportion of people in this sample that succeed is 0.64. a)Calculate the test statistic (z). Give your answer to 2 decimal places. z = b)Calculate the P-value. Give your answer as a decimal to 4 decimal places. P-value =

Answer 1

a)

Test statistics

z = ( - p) / sqrt [ p( 1 - p) / n ]

= (0.64 - 0.60) / sqrt [ 0.60 * ( 1 - 0.60) / 101 ]

= 0.82

b)

p-value = P(Z > z)

= P(Z > 0.82)

= 0.2061 (From Z table)