The customers at a bank complained about long lines and the time they had to spend waiting for service. It is known that the customers at this bank had to wait 7 minutes, on average, before being served. The management made some changes to reduce the waiting time for its customers. A sample of 57 customers taken after these changes were made produced a mean waiting time of 6.5 minutes with a standard deviation of 2.1 minutes. Using this sample mean, the bank manager displayed a huge banner inside the bank mentioning that the mean waiting time for customers has been reduced by new changes. Do you think the bank manager’s claim is justifiable? Use a 10% significance level to answer this question. Use both approaches.

And since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis for this case and makes sense conclude that the true mean is less than 7 minutes

P value

The p value for this case would be given by:

[tex]p_v =P(t_{(56)}<-1.798)=0.0388[/tex]

Since the p value is lower than the significance level provided of 0.1 we have enough evidence to conclude that the true mean is significantly lower than 7 minutes

Step-by-step explanation:

Information provided

[tex]\bar X=6.5[/tex] represent the sample mean for tje waiting times

[tex]s=2.1[/tex] represent the sample standard deviation

[tex]n=57[/tex] sample size

[tex]\mu_o =7[/tex] represent the value to verify

[tex]\alpha=0.1[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value

System of hypothesis

We want to verify if the true mean for this case is less than 7 minutes, the system of hypothesis for this case sre:

Null hypothesis:[tex]\mu \geq 7[/tex]

Alternative hypothesis:[tex]\mu < 7[/tex]

Since we don;t know the population deviation the statistic for the t test is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)

If we replace the info given we got:

[tex]t=\frac{6.5-7}{\frac{2.1}{\sqrt{57}}}=-1.798[/tex]

Critical value

The degreed of freedom are given by:

[tex]df=n-1=57-1=56[/tex]

We are looking for a critical value in the t distribution with 56 degrees of freedom who accumulates 0.10 of the area in the left and we got:

[tex] t_{\alpha}= -1.297[/tex]

And since the calculated value is lower than the critical value we have enough evidence to reject the null hypothesis for this case and makes sense conclude that the true mean is less than 7 minutes

P value

The p value for this case would be given by:

[tex]p_v =P(t_{(56)}<-1.798)=0.0388[/tex]

Since the p value is lower than the significance level provided of 0.1 we have enough evidence to conclude that the true mean is significantly lower than 7 minutes

codiepienagoya codiepienagoya · Answer 2 · 2021-12-08T19:25:47+01:00

Following are the solution to the given question:

Given:

size [tex](n) = 58[/tex]

mean [tex]\mu = 7.4 \ minutes[/tex]

standard deviation [tex]\sigma = 2.3 \ minutes[/tex]

null and alternative hypotheses:

[tex]H_0 : \mu = 8 \ minutes\\\\H_a : \mu < 8 \ minutes\\\\[/tex]

Calculating the test statistic:

So, the observed value

Calculating the degrees of freedom:

[tex]\to 58 - 1 = 57[/tex]

Calculating the range of p-value:

Therefore the t is the critical value at the significance level that is 0.05 with of freedom is,

Therefore,

[tex]\to test \ statistic = -1.987 < -1.672[/tex] rejecting the null hypothesis.

Therefore, the conclusion of the manager's claim is "True".

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