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Virtual call centers are staffed by individuals working out of their homes. An airline is considering employing home agents, but only if their level of customer satisfaction is greater than $80\%$. A test was conducted with home servicing agents. In a sample of size 300 customers, 252 reported that they were satisfied with the service

A. The airline set up the test: $$ \begin{align*} H_0 &: p \leq 0.8 \\ H_1 &: p > 0.8 \end{align*} $$ at level $\alpha = 0.05$ ; p is the true proportion among all customers who will be satisfied by service from home agent

Set up and sun this test, using significance level $$ \alpha =.05 $$. What conclusion do you reach?

$$ \begin{align*} n = 300&, N =252 \\ H_0 &: p \leq 0.8 \\ H_1 &: p > 0.8 \end{align*} $$

Test statistic: One Group Proportion

$$Z = \frac{.84 - .8}{\sqrt{.8(1-.84)/300}} $$

From here I would compare it with the Z value at .05?

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Compare your Z statistic to the Z critical value for a 5% significance. This Z critical value is 1.644 and your Z statistic is 1.732.. Since this is right tailed you compare whether your Z statistic is greater than Z critical. In this case it is, Since it is observed that z = 1.732 > z* = 1.64, it is then concluded that the null hypothesis is rejected...Additionally, The p-value is p = 0.0416, and since p = 0.0416 < .05, it is concluded that the null hypothesis is rejected.. Therefore, there is enough evidence to claim that the population proportion p is greater than po, at the .05 significance level... I'm on mobile so I can't format this nicely but hopefully that helps. In addition the denominator should be the null value (.8)

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  • $\begingroup$ Which formula did you use to calculate the Z statistic 1.732? I used a one group proportions test. $\endgroup$ – user Dec 18 '16 at 23:20
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    $\begingroup$ Yes you have it right. But the denominator is the Null proportion (.8) not .84... so Square root ((.8(1-.8)/300)) $\endgroup$ – Brandon Dec 19 '16 at 1:15

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