Answer :
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.
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