Answer to Question #173595 in Statistics and Probability for ANJU JAYACHANDRAN

Question

7. a) At a call centre, callers have to wait till an operator is ready to take their call. To

monitor this process, 5 calls were recorded every hour for the 8-hour working day.

The data below shows the waiting time in weeks:

Time Sample number

1 2 3 4 5

9 a.m. 8 9 15 4 11

10 7 10 7 6 8

11 11 12 10 9 10

12 12 8 6 9 12

1 p.m. 11 10 6 14 11

2 7 7 10 4 11

3 10 7 4 10 10

4 8 11 11 11 7

Compute the control limits for x and R -charts. Draw the control charts for x and

,R and comments.

Answer 1

Given $\mu_0=26, \bar{x}=20, s=10, n=18.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=26$

$H_1:\mu\not=26$

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test with degrees of freedom $df=n-1=18-1=17$ is $t_c=2.109816.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.109816\}.$

  

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu_0}{s/\sqrt{n}}=\dfrac{20-26}{10/\sqrt{18}}=-2.545584$$

Since it is observed that $|t|=2.545584>2.109816=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $26\ cm,$ at the $0.05$ significance level.

Using the P-value approach:

The p-value for two-tailed, $df=17, \alpha=0.05, t=-2.545584$ is $p=0.020881,$ and since $p=0.020881<0.05=\alpha,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is different than $26\ cm,$ at the $0.05$ significance level.

Range mean:

$\bar{R}=\frac{\sum R}{n}$

$\bar{R}=\frac{9.66}{25}$

$\bar{R}=0.386$

Control limits for mean chart:

$UCL_{\bar{x}} = \bar{\bar{x}} +A_{2}\bar{R}$

 = 20.677 + (0.419)(0.386)

 = 20.83

$CL_{\bar{x}} =\bar{\bar{x}} = 20.677$

$LCL_{\bar{x}} =\bar{\bar{x}} - A_{2}\bar{R}$

$LCL_{\bar{x}} =20.677-(0.419)(0.386)$

$LCL_{\bar{x}} =20.51$

Control limits for range chart:

$LCL_{\bar{R}} =D_{3}\bar{R} = 0.076*0.386 = 0.029$

$CL_{\bar{R}} =0.386$

$UCL_{\bar{R}} =D_{4}\bar{R} = 1.924\times 0.386 = 0.74$

Mean chart is-

Range chart-