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The bank manager wants to show that the new system reduces typical customer waiting times to less than 6 minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6. Letting this mean be µ, in this exercise we wish to investigate whether the sample of 107 waiting times provides evidence to support the claim that µ is less than 6.



For the sake of argument, we will begin by assuming that µ equals 6, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that µ is less than 6. Recall that the mean of the sample of 107 waiting times is = 5.27 and assume that σ, the standard deviation of the population of all customer waiting times, is known to be 2.23.



(a) Consider the population of all possible sample means obtained from random samples of 107 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of ?







(b) Find the mean and standard deviation of the population of all possible sample means when we assume that µ equals 6. (Round your answer to 4 decimal places.)







(c) The sample mean that we have actually observed is = 5.27. Assuming that µ equals 6, find the probability of observing a sample mean that is less than or equal to = 5.27. (Round "z-value" to 2 decimals and final answer to 4 decimal places.)







(d) If µ equals 6, what percentage of all possible sample means are less than or equal to 5.27? What do you conclude about whether the new system has reduced the typical customer waiting time to less than 6 minutes? (Round your answer to 2 decimal places.)



Answer :

The shape of the population sample means is large, the mean and standard deviation is 6 and 0.2155, the probability of sample mean is 0.37172 and it is conclude that mean is less than 6.

Given that sample of 107 waiting times provides evidence to support the claim that µ is less than 6 and for the waiting times mean is [tex]\bar{x}[/tex]=5.27 and standard deviation is 2.23.

(a) It is observed that the sample size n=107, population mean μ=6, sample mean [tex]\bar{x}[/tex]=5.27, and population standard deviation σ=2.23.

The Central Limit Theorem (CLT) states that for a large number of samples, the sample mean tends to approximate the standard value. Using this, it can be asserted that the sample mean follows an approximate normal distribution with μ and variance σ²/n. So, this data will follows a normal distribution because it is very large.

(b) The given mean of all samples is μ=6.

The standard error (SE) is same as the standard deviation of the sample mean. SE is computed as shown below:

SE=√(σ²/n)

SE=σ/√n

SE=2.23/√107

SE=0.2155

In notations,

[tex]\bar{x}\sim N(\mu=6,SE=0.2155)[/tex]

(c) For a single mean (n=1), the z-score is given as follows:

[tex]\begin{aligned}z&=\frac{\bar{x}-\mu}{\sigma}\sim N(0,1)\\ &=\frac{\bar{x}-6}{2.23}\end{aligned}[/tex]

Then, the probability less than or equal to 5.27 is computed as given below:

[tex]\begin{aligned}P(\bar{x}\leq 5.27)&=P\left(\frac{\bar{x}-6}{2.23}\leq \frac{5.27-6}{2.23}\right)\\ &=P(z\leq -0.3273)\\ &=P(z > 0.3273)\\ &=1-P(z \leq 0.3273)\end{aligned}[/tex]

Using the normal table, it is found out that  P(z≤0.3273)=0.6282

[tex]\begin{aligned}P(\bar{x}\leq 5.27)&=1-0.62828\\&=0.37172\end[/tex]

d. If the population mean =6 and sample mean is 5.27 then there is 3.71% that sample mean is less than 5.27 hence we conclude that means is less than 6.

Hence, when  the mean of the sample of 107 waiting times is = 5.27 and assume that σ, the standard deviation is known to be 2.23. the shape of the population sample means is large, the mean and standard deviation is 6 and 0.2155, the probability of sample mean is 0.37172 and it is conclude that mean is less than 6.

Learn more about the standard deviation from here brainly.com/question/16903717

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