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A consulting agency was asked by a large insurance company to investigate if business majors were better salespersons than those with other majors. A sample of 20 salespersons with a business degree showed that they sold an average of 11 insurance policies per week. Another sample of 25 salespersons with a degree other than business showed that they sold an average of 9 insurance policies per week. Assume that the two populations are approximately normally distributed with population standard deviations of \(1.80\) and \(1.35\) policies per week, respectively. a. Construct a \(99 \%\) confidence interval for the difference between the two population means. b. Using a \(1 \%\) significance level, can you conclude that persons with a business degree are better salespersons than those who have a degree in another area?

Short Answer

Expert verified
Without actual calculation, the specific confidence interval and conclusion can't be provided. Nonetheless, the confidence interval will tell us whether there could be a significant difference between the means of the two populations. The hypothesis test will help us decide, at a \(1 \% \) level of significance, if we can be certain that salespersons with a business degree sell more policies on average than those with other degrees.

Step by step solution

01

Identify Parameters

Identify and list the parameters. For the ‘Business Degree’ group, the sample size \(n_1 = 20\), sample mean \(\overline{x}_1 = 11\), and population standard deviation \(σ_1 = 1.80\). For the ‘Other Degree’ group, the sample size \(n_2 = 25\), sample mean \(\overline{x}_2 = 9\), and population standard deviation \(σ_2 = 1.35\). The confidence level for the CI is given as \(99 \% \), and the significance level for the hypothesis test is \(1 \% \).
02

Calculate the Standard Error

Calculate the standard error of the difference between the two sample means. This can be done using the formula: \(SE = \sqrt{ (σ_1^2/n_1) + (σ_2^2/n_2)}\). Substituting the given values we find, \(SE = \sqrt{ (1.80^2/20) + (1.35^2/25)}\).
03

Construct the Confidence Interval

The confidence interval is calculated as \(\overline{x}_1 - \overline{x}_2 \pm Z_{α/2} × SE\), where \(Z_{α/2}\) is the Z value from the standard normal distribution corresponding to the half the significance level (from the z-table). For a \(99\%\) confidence interval, \(Z_{α/2}\) corresponds to \(2.576\). So, CI = 11 - 9 ± 2.576 × SE.
04

Interpret the Confidence Interval

If the calculated Confidence Interval contains only positive numbers, then there is a significant difference between the means of the two populations, therefore, business majors sell more. If the CI contains 0 or negative numbers, there is no significant difference.
05

Hypothesis Testing

Establish the null hypothesis (H0: \(\overline{x}_1 - \overline{x}_2 \leq 0\)) and alternative hypothesis (H1: \(\overline{x}_1 - \overline{x}_2 > 0\)). The test statistic Z is calculated as \((\overline{x}_1 - \overline{x}_2) / SE\). Compare this value with the critical z value for \(1\%\) level of significance (\(2.33\) for a one-tailed test). Reject the null hypothesis if calculated Z > \(2.33\).
06

Conclusions

If the null hypothesis is rejected, there's evidence to suggest that salespersons with a business degree sell more insurance policies on average than those with other degrees. If the null hypothesis is not rejected, then there's no evidence for this.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used to decide if a certain claim about a population parameter is supported by evidence from a sample. In this case, we're testing if business majors are better salespersons than those with different degrees.
The **null hypothesis (H0)** states that there is no difference in sales performance between the two groups, or that salespersons with business degrees are not better (i.e., \(ar{x}_1 - ar{x}_2 \leq 0\)). The **alternative hypothesis (H1)** suggests the opposite, indicating that business majors are indeed better (i.e., \(ar{x}_1 - ar{x}_2 > 0\)).
To test these hypotheses, a test statistic is calculated and compared against a critical value. If this test statistic is greater than the critical value, the null hypothesis is rejected in favor of the alternative hypothesis, indicating meaningful evidence. In this example, the critical value comes from a standard normal distribution, reflecting a significance level of 1%.
Population Means
Population means represent the average of a particular characteristic across all individuals in a population. We often use sample means as estimates since it's usually impractical to measure an entire population. In this example, we're comparing the mean number of insurance policies sold by two groups: one with business degrees and one without.
  • The **sample mean for business degree holders** is \(\bar{x}_1 = 11\) policies per week.
  • The **sample mean for other degree holders** is \(\bar{x}_2 = 9\) policies per week.
These means form the basis for our hypothesis test and the confidence interval assessed. By evaluating these samples, we aim to infer differences between the entire groups they represent in the broader population.
Standard Error
Standard error (SE) measures the dispersion of sample means around the population mean. It provides insight into how much the sample mean is expected to vary, which is crucial in hypothesis testing and constructing confidence intervals.
For this example, the **standard error of the difference** between the two sample means is calculated using the formula:
\[ SE = \sqrt{ \left(\frac{\sigma_1^2}{n_1}\right) + \left(\frac{\sigma_2^2}{n_2}\right) } \] where \(\sigma_1\) and \(\sigma_2\) are the population standard deviations, and \(n_1\) and \(n_2\) are the sample sizes for each group. A smaller SE indicates more precise estimate of the true population mean difference, enabling more confident inferences.
Significance Level
The significance level, often denoted as \(\alpha\), represents the threshold for rejecting the null hypothesis. It indicates the probability of committing a Type I error, which means rejecting a true null hypothesis. In this exercise, a significance level of 1% is used.
A lower significance level means we require stronger evidence against the null hypothesis to consider the result statistically significant. This reduces the risk of mistakenly concluding that one group is better than another when there truly is no difference. The choice of significance level affects the critical value used in hypothesis testing. For a 1% level, the critical value for a one-tailed test is 2.33.
This ensures that only results with compelling evidence are considered significant, providing a rigorous standard for validating the findings in this sales performance analysis.

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Most popular questions from this chapter

According to a Bureau of Labor Statistics report released on March 25, 2015 , statisticians earn an average of \(\$ 84,010\) a year and accountants and auditors earn an average of \(\$ 73,670\) a year (www.bls. gov). Suppose that these estimates are based on random samples of 2000 statisticians and 1800 accountants and auditors. Further assume that the sample standard deviations of the annual earnings of these two groups are \(\$ 15,200\) and \(\$ 14,500\), respectively, and the population standard deviations are unknown and unequal for the two groups. a. Construct a \(98 \%\) confidence interval for the difference in the mean annual earnings of the two groups, statisticians and accountants and auditors. b. Using a \(1 \%\) significance level, can you conclude that the average annual earnings of statisticians is higher than that of accountants and auditors?

Refer to the previous exercise. Suppose Gamma Corporation decides to test governors on seven cars. However, the management is afraid that the speed limit imposed by the governors will reduce the number of contacts the salespersons can make each day. Thus, both the fuel consumption and the number of contacts made are recorded for each car/salesperson for each week of the testing period, both before and after the installation of governors. Suppose that as a statistical analyst with the company, you are directed to prepare a brief report that includes statistical analysis andinterpretation of the data. Management will use your report to help decide whether or not to install governors on all salespersons' cars. Use \(90 \%\) confidence intervals and \(.05\) significance levels for any hypothesis tests to make suggestions. Assume that the differences in fuel consumption and the differences in the number of contacts are both normally distributed.

The following information was obtained from two independent samples selected from two normally distributed populations with unknown but equal standard deviations. $$ \begin{array}{lll} n_{1}=21 & \bar{x}_{1}=13.97 & s_{1}=3.78 \\ n_{2}=20 & \bar{x}_{2}=15.55 & s_{2}=3.26 \end{array} $$ a. What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b Construct a \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

An economist was interested in studying the impact of the recession of a few years ago on dining out, including drive-through meals at fast-food restaurants. A random sample of 48 families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week indicated that they reduced their spending on dining out by an average of \(\$ 31.47\) per week, with a sample standard deviation of \(\$ 10.95 .\) Another random sample of 42 families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week reduced their spending on dining out by an average of \(\$ 35.28\) per week, with a sample standard deviation of \(\$ 12.37\). (Note that the two groups of families are differentiated by the number of family members.) Assume that the distributions of reductions in weekly dining-out spendings for the two groups have unknown and unequal population standard deviations. a. Construct a \(90 \%\) confidence interval for the difference in the mean weekly reduction in dining out spending levels for the two populations. b. Using a \(5 \%\) significance level, can you conclude that the average weekly spending reduction for all families of four with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week is less than the average weekly spending reduction for all families of five with discretionary incomes between \(\$ 300\) and \(\$ 400\) per week?

A state that requires periodic emission tests of cars operates two emission test stations, \(\mathrm{A}\) and \(\mathrm{B}\), in one of its towns. Car owners have complained of lack of uniformity of procedures at the two stations, resulting in different failure rates. A sample of 400 cars at Station A showed that 53 of those failed the test; a sample of 470 cars at Station B found that 51 of those failed the test. a. What is the point estimate of the difference between the two population proportions? b. Construct a \(95 \%\) confidence interval for the difference between the two population proportions. c. Testing at a \(5 \%\) significance level, can you conclude that the two population proportions are different? Use both the critical-value and the \(p\) -value approaches.

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