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The management at New Century Bank claims that the mean waiting time for all customers at its branches is less than that at the Public Bank, which is its main competitor. A business consulting firm took a sample of 200 customers from the New Century Bank and found that they waited an average of \(4.5\) minutes before being served. Another sample of 300 customers taken from the Public Bank showed that these customers waited an average of \(4.75\) minutes before being served. Assume that the standard deviations for the two populations are \(1.2\) and \(1.5\) minutes, respectively. a. Make a \(97 \%\) confidence interval for the difference between the two population means. b. Test at the \(2.5 \%\) significance level whether the claim of the management of the New Century Bank is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.01 ?\) What if \(\alpha=.05 ?\)

Short Answer

Expert verified
To briefly give an answer without going through all the calculations, we would first construct a 97% confidence interval for the difference in mean waiting times between the two banks. Then we would perform a hypothesis test at the 2.5% significance level to test the management's claim. Lastly, we would calculate the p-value and use it to decide whether to reject the null hypothesis at the 0.01 and 0.05 significance levels.

Step by step solution

01

Create a 97% Confidence Interval

A 97% confidence interval for the difference between the two population means is given by:\[(\bar{x}-\bar{y})\pm t_{\alpha/2}(s_{\bar{x}-\bar{y}})\]In this equation, \(\bar{x}\) and \(\bar{y}\) are the sample means of the New Century Bank and the Public Bank respectively. The term \(s_{\bar{x}-\bar{y}}\) is the standard deviation of the difference between the sample means and \(t_{\alpha/2}\) is the t-score associated with the desired level of confidence (97%). The sample means and standard deviations are given in the problem. The t-score can be found using a table or statistical software. The resulting interval will give us the range of values that the true difference between the two population means could possibly fall into with 97% confidence.
02

Perform a Hypothesis Test

The null hypothesis (\(H_0\)) is that the mean waiting time at New Century Bank is equal to the mean waiting time at the Public Bank. The alternative hypothesis (\(H_A\)) is that the mean waiting time at New Century Bank is less than the mean waiting time at the Public Bank. We use a t-test to compare the sample means and judge whether the observed difference is statistically significant. If the test statistic falls into the rejection region (determined by the 2.5% significance level), we reject the null hypothesis in favour of the alternative.
03

Calculate the p-value

The p-value is the probability, if the null hypothesis is true, of observing a test statistic as extreme as, or more extreme than, the observed test statistic. We calculate it using the observed test statistic and the t-distribution. If the calculated p-value is less than a given significance level (\(\alpha\)), we reject the null hypothesis.
04

Interpret the p-value

If the p-value is less than 0.01, then we would reject the null hypothesis if \(\alpha=.01\). If the p-value is less than 0.05, we would also reject the null hypothesis if \(\alpha=.05\). Otherwise, we fail to reject the null hypothesis. This decision rule helps us to determine whether the evidence is strong enough to support the management's claim at the desired significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval is a range of values that estimates an unknown population parameter. In our exercise, a 97% confidence interval was created for the difference in waiting times between New Century Bank and Public Bank. This interval gives us a range within which the true difference between the two population means is likely to fall, with 97% confidence.

To calculate a confidence interval, we need the sample means, the standard deviations of both populations, and the size of each sample. These values are plugged into the formula for the confidence interval. We also use a t-score, which corresponds to our confidence level. The t-score adjusts for the fact that we are estimating the population standard deviation based on the sample. Once our calculation is complete, we are left with a range. If this range does not include zero, it could suggest a significant difference between the two population means.
p-value
The p-value is a key concept in hypothesis testing that helps us understand the strength of our results. It represents the probability of obtaining test results as extreme as the observed results, assuming the null hypothesis is true. In simpler terms, it shows how likely the sample data is under the assumption of no effect or no difference.

In the context of our problem, the p-value was calculated to determine if the difference in waiting times is statistically significant. A low p-value typically indicates that the observed data is unlikely to occur if the null hypothesis is true, leading us to consider rejecting it. For this exercise, if the p-value is less than the significance level (2.5% significance level was used), we reject the null hypothesis in favor of the alternative. This would support the claim that the New Century Bank's average waiting time is indeed shorter than its competitor's.
Null Hypothesis
The null hypothesis (\(H_0\)) is a default statement or position that there is no effect or no difference. In our specific testing scenario, the null hypothesis states that there is no difference in the mean waiting times between customers at New Century Bank and Public Bank.

Testing the null hypothesis starts with assuming it is true, which means any differences between the sample means are attributed to random sampling variability. Through statistical tests, we analyze whether our data provides enough evidence to reject this assumption. If the evidence is strong (e.g., a low p-value), we consider rejecting \(H_0\) in favor of the alternative hypothesis, which posits a difference.
t-test
A t-test is a statistical test used to compare the means from two groups to determine if they are significantly different. In the context of this problem, a t-test was applied to compare the mean waiting times of customers between New Century Bank and Public Bank.

To conduct a t-test, we calculate the t-statistic, which measures the size of the difference relative to the variation in our sample data. This statistic depends on the difference between the sample means, the standard deviations, and the number of observations in each sample. The result is compared against a t-distribution to determine significance. If the calculated t-statistic exceeds the critical t-value (defined by our significance level), the null hypothesis is rejected, supporting the conclusion that one mean is indeed different from the other.

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Most popular questions from this chapter

A company claims that its medicine, Brand \(A\), provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. $$ \begin{array}{cccc} \hline & & \text { Mean of } & \text { Standard Deviation } \\ \text { Brand } & \text { Sample Size } & \text { Relief Times } & \text { of Relief Times } \\ \hline \text { A } & 25 & 44 & 11 \\ \text { B } & 22 & 49 & 9 \\ \hline \end{array} $$ a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at the \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B.

A local college cafeteria has a self-service soft ice cream machine. The cafeteria provides bowls that can hold up to 16 ounces of ice cream. The food service manager is interested in comparing the average amount of ice cream dispensed by male students to the average amount dispensed by female students. A measurement device was placed on the ice cream machine to determine the amounts dispensed. Random samples of 85 male and 78 female students who got ice cream were selected. The sample averages were \(7.23\) and \(6.49\) ounces for the male and female students, respectively. Assume that the population standard deviations are \(1.22\) and \(1.17\) ounces, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of ice cream amounts dispensed by all male and female students at this college, respectively. What is the point estimate of \(\mu_{1}-\mu_{2} ?\) b. Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). c. Using the \(1 \%\) significance level, can you conclude that the average amount of ice cream dispensed by male college students is larger than the average amount dispensed by female college students? Use both approaches to make this test.

A June 2009 Gallup Poll asked a sample of Americans whether they trust specific groups or individuals when it comes to making recommendations about healthcare reform. Sixty percent of Democrats and \(68 \%\) of Republicans stated that they trust doctors' opinions about healthcare reform (Source: http://www.gallup.com/poll/120890/Healthcare-Americans-Trust-Physicians- Politicians.aspx). Suppose this survey included 340 Democrats and 306 Republicans. a. Make a \(90 \%\) confidence interval for the difference in the population proportions for the two groups of people. b. At the \(5 \%\) significance level, can you conclude that the proportion of all Democrats who trust doctors' opinions about healthcare reform differs from the proportion of all Republicans who trust doctors' opinions about healthcare reform?

A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours.

Assuming that the two populations are normally distributed with unequal and unknown population standard deviations, construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) for the following. $$ \begin{array}{lll} n_{1}=14 & \bar{x}_{1}=109.43 & s_{1}=2.26 \\ n_{2}=15 & \bar{x}_{2}=113.88 & s_{2}=5.84 \end{array} $$

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