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According to a report in The New York Times, in the United States, accountants and auditors earn an average of \(\$ 70,130\) a year and loan officers carn \(\$ 67,960\) a year (Jessica Silver-Greenberg, The New York Times, April 22,2012 ). Suppose that these estimates are based on random samples of 1650 accountants and auditors and 1820 loan officers. Further assume that the sample standard deviations of the salaries of the two groups are \(\$ 14,400\) and \(\$ 13,600\), respectively, and the population standard deviations are equal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean salaries of the two groupsaccountants and auditors, and loan officers. b. Using a \(1 \%\) significance level, can you conclude that the average salary of accountants and auditors is higher than that of loan officers?

Short Answer

Expert verified
Without the exact calculations, it's difficult to give a specific answer. But the logic should be: If the calculated z-score for the difference in mean salaries (calculated in Step 4) exceeds the critical z-score for a 1% significance level, it can be concluded that the average salary of accountants and auditors is higher than that of loan officers at a 1% significance level. The exact numbers for the confidence interval will depend on the computed standard error (Step 2) and the corresponding z-score for a 98% confidence level (Step 3).

Step by step solution

01

Calculate the difference in the mean salaries

Firstly, calculate the difference in the mean salaries of accountants and auditors and loan officers. This is done by subtracting the mean salary of the loan officers from that of the accountants and auditors: \( \mu1 - \mu2 = \$70,130 - \$67,960 = \$2,170\)
02

Calculate the standard error of the mean difference

The standard error of the mean difference can be estimated using the formula for the standard error of the difference between two means. We also have to assume that population standard deviations are equal as mentioned:\( SE = \sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} } \)Where \( s1 = \$14,400 \), \( s2 = \$13,600 \), \( n1 = 1650 \) and \( n2 = 1820 \).After plugging these values into the formula, calculate the standard error.
03

Construct the confidence interval

Now we're ready to construct the 98% confidence interval for the difference in means. The formula for this is:\( CI = (\mu1 - \mu2) \pm z \times SE \)Where \( z \) is the z-score that corresponds to the desired confidence level (98%). You can look up this value in a standard normal distribution table or use a z-score calculator. Calculate the confidence interval using these values.
04

Conduct the hypothesis test

Next, the hypothesis test at a 1% significance level can be conducted. The null hypothesis ( \( H_0 \) ) would be that there is no difference in the mean salaries of accountants and auditors and loan officers, and the alternative hypothesis ( \( H_1 \) ) would be that accountants and auditors earn more. If the calculated z-score exceeds the critical z-score for a 1% significance level, then we reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.

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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 provides us with a range of values within which the true difference in mean salaries between two groups is expected to lie. In our example, this includes the average salaries between accountants and auditors versus loan officers. The aim is to understand how much the average salary differs between these two professions, based on sample data. We express this difference using a specific percentage, known as the confidence level. For instance, when we compute a 98% confidence interval, we can be 98% sure that the true mean difference falls within the range calculated.

The formula to calculate the confidence interval for the difference in means is:
  • CI = (mean difference) ± (z-score * standard error)
This formula helps us determine the interval by considering both the mean difference observed and the variation expected from sampling errors, indicated by the z-score and standard error.

In practice, a confidence interval gives us a tangible way to express uncertainty and variability of the data collected from samples. It essentially allows us to infer about the population means and their differences with a specified level of confidence.
Mean Difference
The mean difference is a vital statistic, especially in comparative studies. It represents the gap between the average values from two distinct groups or samples. In our context, the mean difference tells us how much more, or less, accountants and auditors earn compared to loan officers.

To calculate this, you simply subtract the average salary of one group from the other. Here,
  • Mean Difference = Average Salary of Accountants - Average Salary of Loan Officers = 70,130 - 67,960 = 2,170
This value states that, on average, accountants and auditors earn $2,170 more than loan officers.

Understanding the mean difference is key to interpreting the results of any comparative analysis. It not only highlights the magnitude of variation between groups, but also serves as a basis to assess the impact of differences observed. Whether this difference is considered significant or not often depends on further statistical testing, such as confidence intervals or hypothesis testing, carried out in conjunction with mean difference calculations.
Z-score
A z-score is essentially a measure of how many standard deviations an element is from the mean. When conducting hypothesis testing or forming confidence intervals, the z-score is crucial for determining the position of a data point respecting our population standard.

In our example concerning salaries, the z-score helps us decide how extreme or typical a mean difference is, considering the natural variation (standard error) inherent in sampling. To find the z-score that corresponds to a specific confidence level, like 98%, tables or statistical software systems provide these values based on standard normal distribution.

Using a z-score transforms raw data into a standardized form, which allows for comparison across different groups and scenarios. Specifically, in hypothesis testing, it also assists in comparing the calculated statistics against critical values to determine whether the null hypothesis can be rejected or not.
Standard Error
The standard error is a statistical metric which measures how precise a sample mean approximates the actual population mean. It is especially applicable when sampling is involved since it accounts for sampling variability. In our case, it quantifies the uncertainty in mean difference estimation between accountants' and loan officers' salaries.

The formula to calculate the standard error of the mean difference is:
  • SE = \[\sqrt{ \frac{s1^2}{n1} + \frac{s2^2}{n2} }\]
This involves parameters such as the sample standard deviations (\(s1\), \(s2\)) and the respective sample sizes (\(n1\), \(n2\)).

Smaller standard errors indicate that the sample mean is a more accurate reflection of the population mean, while a larger standard error implies more spread. Essentially, the lower the standard error, the more reliable our estimates are, aiding us in constructing tighter, more informative confidence intervals.

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Most popular questions from this chapter

A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. A. Construct a \(98 \%\) confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using a \(1 \%\) significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store B?

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 population means. b. Test at a 2.5\% significance level whether the claim of the management of the New Century Bank is truc. 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 \(\alpha=05\) ?

The standard recommendation for automobile oil changes is once every 3000 miles. A local mechanic is interested in determining whether people who drive more expensive cars are more likely to follow the recommendation. Independent random samples of 45 customers who drive luxury cars and 40 customers who drive compact lower-price cars were selected. The average distance driven between oil changes was 3187 miles for the luxury car owners and 3214 miles for the compact lower-price cars. The sample standard deviations were \(42.40\) and \(50.70\) miles for the luxury and compact groups, respectively. Assume that the population distributions of the distances between oil changes have the same standard deviation for the two populations. a. Construct a \(95 \%\) confidence interval for the difference in the mean distances between oil changes for all luxury cars and all compact lower-price cars. b. Using a \(1 \%\) significance level, can you conclude that the mean distance between oil changes is less for all luxury cars than that for all compact lower-price cars?

A high school counselor wanted to know if tenth-graders at her high school tend to have more free time than the twelfth-graders. She took random samples of 25 tenth-graders and 23 twelfth-graders. Each student was asked to record the amount of free time he or she had in a typical week. The mean for the tenthgraders was found to be 29 hours of free time per week with a standard deviation of \(7.0\) hours. For the twelfthgraders, the mean was 22 hours of free time per week with a standard deviation of \(6.2\) hours. Assume that the two populations are normally distributed with equal but unknown population standard deviations. a. Make a \(90 \%\) confidence interval for the difference between the corresponding population means. b. Test at a \(5 \%\) significance level whether the two population means are different.

Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8 -ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8-ounce low-fat yogurt cups produced by the two companies. b. Test at a \(1 \%\) significance level whether Maine Mountain Dairy's claim 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 \(\alpha=05\) ? What if \(\alpha=.025\) ?

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