Answer :
Answer:
We reject null hypothesis.
Step-by-step explanation:
We are given that Tango Furniture sells furniture online, arriving at customers homes needing to be assembled. Suppose 3.6% of their Viridian chairs arrive without all the parts.
In trying to improve the system, Hakim reorganizes the packaging system of the Viridian and after testing 500 chairs, only 1% have missing parts.
We have to test is this result a statistically significant improvement.
Let p = % of Viridian chairs arrive without all the parts
SO, Null Hypothesis, [tex]H_0[/tex] : p [tex]\geq[/tex] 3.6% {means that % of Viridian chairs arrive without all the parts is greater than or equal to 3.6%}
Alternate Hypothesis, [tex]H_a[/tex] : p < 3.6% {means that % of Viridian chairs arrive without all the parts is less than 3.6%}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = % of Viridian chairs arrive without all the parts in a testing of
500 chairs = 1%
n = sample of chairs = 500
So, test statistics = [tex]\frac{0.01-0.036}{\sqrt{\frac{0.01(1- 0.01)}{500} } }[/tex]
= -5.843
Since in the question we are not given with the significance level so we assume it to be 5%. So, at 0.05 level of significance, the z table gives critical value of -1.6449 for one-tailed test. Since our test statistics is less than the critical value of z so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.
Therefore, we conclude that % of Viridian chairs arrive without all the parts is less than 3.6% which means that there is statistically significant improvement.