Answer to Question #262519 in Statistics and Probability for Andrei

Question

A quality-control manager at an amusement park feels that the amount of time that people spend waiting in line to ride the roller coaster is too long. To determine if a new loading and unloading procedure is effective in reducing the wait time, she measures the amount of time (in minutes) people are waiting in line for seven days. To make a reasonable comparison, she chooses times when weather conditions are similar.

Mon.   Tues.    Wed.   Thurs.  Fri.   Sat.   Sun

2 pm. 2 pm.   2 pm.   2 pm.   2 pm.  4 pm.  12 noon

Wait time before x1 12. 26. 20. 38. 57. 82  57

Wait time before x2. 11. 28. 19. 36. 59. 75. 55

Is the new loading and unloading procedure effective in reducing the wait time at the α = 0.05 level of significance? 

Answer 1

Our sample is dependent and we should consider using matched/paired t-test for our analysis.

To perform this test, we begin by finding the difference between pairs of observation for each student. We shall use the notation $d,$ to represent the differences and use it to make a decision.

The hypotheses to be tested are,

$H_0:\mu_d=0$

$Against$

$H_1:\mu_d\gt0$, where $d$ is the difference between each pair of observations.

To test these hypotheses, we first determine the mean of the differences$(\bar{d})$ given by,

$\bar{d}=(\sum(d))/n$, where $n=7$

Now,

$\sum(d)=9$

Therefore,

$\bar{d}=9/7=1.285714$

Variance of the differences is given by,

$V(d)=\sum(d-\bar{d})^2/(n-1)$

$V(d)=55.42857/6=9.238095$.

The standard deviation $SD(d)$ is given by,

$SD(d)=\sqrt{V(d)}=\sqrt{9.238095}=3.039324$.

The test statistic is given by,

$t^*=\bar{d}/(SD(d)/\sqrt n)$

$t^*=1.285714/(3.039424/\sqrt {7})$

$t^*=1.285714/1.148794=1.119186.$

$t^*$ is compared with the student's t distribution with $(n-1)=7-1=6$ degrees of freedom at $\alpha=0.05$.

The table value $t_{\alpha,6}=t_{0.05,6}=t_{0.05,6}=1.943$ and reject the null hypothesis if $t^*\gt t_{0.05,6}.$

Since $t^*=1.119186\lt t_{0.05,6}=1.943$, we fail to reject the null hypothesis and conclude that sufficient evidence do not exist to show that the  new loading and unloading procedure is effective in reducing the wait time at 5% level of significance.