Question

$Screenshot_20210209_101224.jpg$

A service counter manager wishes to ensure that an average of 16 seconds to serve a customer. In order to analyse the efficiency of the service process, he takes a random sample of 56 customers. The mean service time of the sample is 15.8 seconds. Assume that the population standard deviation is 0.8 seconds.

(a) State the null and the altemative hypotheses for the test.

(b) Use the p-value approach to test the manager's concem at \(\alpha=0.05\).

(c) Confim your result in (ii) by using the critical value approach.

(d) State the recommendation that the manager can present to the company.

Answer 1

a)

The hypothesis being tested is:

Null Hypothesis: A service counter manager wishes to ensure that an average of 16 seconds to serve a customer that is \(\mathrm{u}=16\)

Alternate Hypothesis: A service counter manager wishes to ensure that an average of 16 seconds to serve a customer is not equal to 16 that is \(u \neq 16\)

This is a two-tailed test.

b)

$$ z=\frac{\bar{X}-\mu_{0}}{\sigma / \sqrt{n}}=\frac{15.8-16}{0.8 / \sqrt{56}}=-1.871 $$

The P value= 0.0614 [Using Z table]

Since P value> alpha. We fail to reject the null hypothesis and there is not sufficient evidence to support the claim that the service counter manager wishes to ensure that an average of 16 seconds to serve a customer is not equal to 16 that is \(u \neq 16\) at \(5 \%\) level of significance.

c)

\(a=0.05\)

The critical value using z table is 1.96 .

$$ z_{c}=1.96 $$

Since it is observed that \(|z|=1.871 \leq z_{c}=1.96,\) it is then concluded that the null hypothesis is not rejected.

d)

There is enough evidence to support the claim that the average time to serve a customer is 16 seconds. Thus, we can conclude that this service process is very efficient.

Answer 2

The information given to us is:

The sample size, \(\mathrm{n}=56\)

The sample mean \(=15.8\) sec

The population standard deviation \(=0.8 \mathrm{sec}\)

The level of significance, \(a=0.05\)

a. To test whether it takes 16 seconds to serve a customer, the null and alternative hypothesis are:

\(\mathrm{H}_{0}: \mu=16,\) against

\(\mathrm{H}_{1}: \mu \neq 16\) (two-tailed)

b. As the sample size is large (i.e, \(n>30\) ) and the population standard deviation is known, the test statistic is given by:

$$ Z=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) $$

Thus,

$$ \begin{aligned} z &=\frac{\bar{x}-\mu}{\sigma / \sqrt{n}} \\ &=\frac{15.8-16}{0.8 / \sqrt{56}} \\ &=-1.8708 \end{aligned} $$

For a 2 -tailed test, the \(p\) -value corresponding to the test statistic is:

p-value \(=2 P(Z<-1.8708)=0.06137\)

The decision rule is, if the p-value is less than the level of significance, reject the null hypothesis.

Since the \(p\) -value \(=0.03069\) is greater than the level of significance \(a=0.05,\) we fail to reject the null hypothesis. there is not sufficient evidence to reject the claim, 'the average time taken to serve a customer is 16 seconds'.

c. Using the critical value approach, the decision rule is:

If \(\left|z_{c a l}\right|>\left|z_{c r i t}\right|,\) we reject the null hypothesis.

Here, at \(a=0.05,\) the critical value of \(z\) are \(\pm 1.96,\) thus \(\left|z_{c r i t}\right|=1.96\)

and the calculated value of \(z\) is \(z_{c a l}=-1.8708,\) thus \(\left|z_{c a}\right|=1.8708\)

Since, \(\left|z_{c a \mid}\right|=1.8708<\left|z_{c r i t}\right|=1.96,\) we fail to reject the null hypothesis. There is not sufficient evidence to

reject the claim, 'the average time taken to serve a customer is 16 seconds'.

d. There is enough evidence to support the claim that average time to serve a customer is 16 seconds. So we can say this service process is very efficient.