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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 0.63.
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 (TT) to above 0.63. A hypothesis test is conducted in order to find out.
The null and alternative hypotheses are:
H0: Pi symbol = 0.63
HA: Pi symbol > 0.63
To run the test, the manager randomly selects a sample of 102 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 0.67. You may find this standard normal table useful throughout this question.
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 =

Sagot :

The test statistic z = 0.84 and P-value is 0.2005 when there is a line of unassembled furniture produced by The Mean Corporation.

Given that,

There is a line of unassembled furniture produced by The Mean Corporation. The customer care manager is aware that 0.63 of customers have had no issues putting together their furniture at home based on past data.

The null and alternative hypotheses are:

H₀: π = 0.63

H₁: π > 0.63

We have to find

(a) Calculate the test statistic (z).

(b) Calculate the P-value.

We know that,

a) z = p - π / √(π(1-π)/n)

      = 0.67 - 0.63 / √(0.63×0.37/102)

      = 0.04 / 0.0478

      = 0.84

b) p value(one tailed) is 0.2005.

Therefore, The test statistic z = 0.84 and P-value is 0.2005 when there is a line of unassembled furniture produced by The Mean Corporation.

To learn more about statistic visit: https://brainly.com/question/11969248

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