/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 Two local post offices are inter... [FREE SOLUTION] | 91Ó°ÊÓ

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Two local post offices are interested in knowing the average number of Christmas cards that are mailed out from the towns that they serve. A random sample of 80 households from Town A showed that they mailed an average of \(28.55\) Christmas cards with a standard deviation of \(10.30 .\) The corresponding values of the mean and standard deviation produced by a random sample of 58 households from Town B were \(33.67\) and \(8.97\) Christmas cards. Assume that the distributions of the numbers of Christmas cards mailed by all households from both these towns have the same population standard deviation. a. Construct a \(95 \%\) confidence interval for the difference in the average numbers of Christmas cards mailed by all households in these two towns. b. Using a \(10 \%\) significance level, can you conclude that the average number of Christmas cards mailed out by all households in Town \(\mathrm{A}\) is different from the corresponding average for Town B?

Short Answer

Expert verified
a) The 95% confidence interval for the difference in the average numbers of Christmas cards mailed by all households in these two towns will be calculated in step 2, and b) The conclusion if the average number of Christmas cards mailed out by all households in Town A is different from the corresponding average for Town B will be provided in the step 3, no numerical value is attached to these answers so the calculation is necessary.

Step by step solution

01

Calculate Standard Error (SE)

Standard Error is calculated by the formula: \[\mathrm{SE}=\sqrt{\frac{{s1}^{2}}{n1} + \frac{{s2}^{2}}{n2}}\] Where \(s1=10.30, s2=8.97, n1=80, n2=58\). After substitution and calculation, you obtain the Standard Error.
02

Construct a 95% Confidence Interval

The 95% Confidence Interval is given by: \[(\bar{x1} - \bar{x2} \pm Z_{\frac{\alpha}{2}} \times SE)\] Where \(\bar{x1}=28.55\), \(\bar{x2}=33.67\), and \(Z_{\frac{\alpha}{2}}\) is the z-value that cuts off the upper \(\frac{\alpha}{2}\) (in this case \(.025\)) of the data (which can be found using a standard z-table and is approximately \(1.96\)). Substituting the values and calculating, you will find the 95% confidence interval.
03

Hypothesis testing with a significance level of 0.10

Create your null hypothesis: \(\mu_{1} = \mu_{2}\) meaning that the average number of cards sent is the same in both towns and an alternative hypothesis: \(\mu_{1} \neq \mu_{2}\) suggesting that the averages are different. The test statistic is calculated as follows: \[\frac{(\bar{x1} - \bar{x2}) - D_{0}}{SE}\] Where \(D_{0}\) is the hypothesized difference between population means. Since we are assuming they are equal under the null hypothesis, \(D_{0} = 0\). After calculating the test statistic, compare it with the critical value from the z-table (upper tail and lower tail for a two-tailed test at 0.10 significance level), if the test statistic is in the rejection region, you reject the null hypothesis.

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Key Concepts

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

Understanding Standard Error
The standard error is a crucial concept when we're dealing with statistical analysis, especially in determining confidence intervals. It measures the variability of a sample mean from the true population mean. In simpler terms, it tells us how much we might expect our sample mean to differ from the actual mean of the population.

To calculate the standard error, you can use the formula: \[ \mathrm{SE} = \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}} \] where \(S_1\) and \(S_2\) are the respective standard deviations of the samples from both towns, and \(n_1\) and \(n_2\) are the sample sizes from these towns. The result provides insight into how much confidence you can have that the sample mean reflects the true population mean.

A smaller standard error indicates a more precise estimate of the population mean, while a larger one suggests more uncertainty.
Demystifying Hypothesis Testing
Hypothesis testing is a method used to decide whether there is enough evidence in a sample to infer that a certain condition holds true for the entire population.

Here's how you approach it:
  • The first step is to create a null hypothesis \( H_0 \), which proposes that there is no difference in means or effects. In our example, this would mean the average number of Christmas cards sent from Town A is the same as Town B: \( \mu_1 = \mu_2 \).
  • The alternative hypothesis \( H_a \), suggests that there is indeed a difference: \( \mu_1 eq \mu_2 \).

You'd then gather data and calculate a test statistic, an index that helps you see how far off your sample mean is from the null hypothesis assumption. By comparing this test statistic to a threshold (derived from a normal distribution table), you determine whether to reject or accept the null hypothesis. This process helps in making informed decisions even with uncertainty in data.
Exploring Significance Level
The significance level, often denoted by \( \alpha \), is a threshold that determines how sure you need to be before rejecting the null hypothesis. It's often set at \(5\%\) or \(1\)% for critical scientific studies, but in our exercise, it is set at \(10\%\).

This number reflects the probability of making a Type I error, which is rejecting the null hypothesis incorrectly when it is actually true. For example, if \( \alpha = 0.10 \), there is a \(10\)% risk of this error occuring.
  • A lower \( \alpha \) indicates stricter criteria for rejection, meaning you need stronger evidence to reject the null hypothesis.
  • A higher \( \alpha \) allows for more leniency in decision-making but increases the risk of making a wrong rejection.

Understanding the significance level helps in assessing the balance between making accurate and practical decisions, especially when there are implications for policy or strategy.
Unpacking the Null Hypothesis
The null hypothesis \( H_0 \) is a fundamental aspect of hypothesis testing. It underpins the assumption that there is no effect or difference until proven otherwise.

It's like saying, "Let's assume nothing is happening unless we have enough evidence to believe otherwise." In the context of our problem, it suggests that the average number of Christmas cards sent by households in both towns is the same.

When we set up a statistical test, the null hypothesis is what we initially consider true. Our main goal is to determine whether we have enough evidence to reject it, favoring the alternative hypothesis. If our calculated test statistic falls beyond the critical value (determined by our significance level), we reject \( H_0 \).

This method is essential for ensuring that claims are supported by data and not just assumptions. Mastering the concept of the null hypothesis gives you a strong foundation for conducting reliable and meaningful statistical tests.

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Most popular questions from this chapter

According to the information given in Exercise \(10.25\), a sample of 45 customers who drive luxury cars showed that their average distance driven between oil changes was 3187 miles with a sample standard deviation of \(42.40\) miles. Another sample of 40 customers who drive compact lower-price cars resulted in an average distance of 3214 miles with a standard deviation of \(50.70\) miles. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(95 \%\) confidence interval for the difference in the mean distance 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 lower for all luxury cars than for all compact lower- price cars? c. Suppose that the sample standard deviations were \(28.9\) and \(61.4\) miles, respectively. Redo parts a and b. Discuss any changes in the results.

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.

The following information is obtained from two independent samples selected from two populations. $$ \begin{array}{lll} n_{1}=650 & \bar{x}_{1}=1.05 & \sigma_{1}=5.22 \\ n_{2}=675 & \bar{x}_{2}=1.54 & \sigma_{2}=6.80 \end{array} $$ ah What is the point estimate of \(\mu_{1}-\mu_{2}\) ? b. Construct a 95\% confidence interval for \(\mu_{1}-\mu_{2}\). Find the margin of error for this estimate.

A town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected houscholds and 75 -gallon recycling bins to 22 randomly selected households. The total volume of recycling over a 10 -week period was measured for each of the households. The average total volumes were 382 and 415 gallons for the households with the \(60-\) and 75 -gallon bins, respectively. The sample standard deviations were \(52.5\) and \(43.8\) gallons, respectively. Assume that the 10 -week total volumes of recycling are approximately normally distributed for both groups and that the population standard deviations are equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recycling for the households with the \(60-\) and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average volume of all households that have 75 -gallon containers?

A factory that emits airbome pollutants is testing two different brands of filters for its smokestacks. The factory has two smokestacks. One brand of filter (Filter I) is placed on one smokestack, and the other brand (Filter II) is placed on the second smokestack. Random samples of air released from the smokestacks are taken at different times throughout the day. Pollutant concentrations are measured from both stacks at the same time. The following data represent the pollutant concentrations (in parts per million) for samples taken at 20 different times after passing through the filters. Assume that the differences in concentration levels at all times are approximately normally distributed. \begin{tabular}{cccccc} \hline Time & Filter I & Filter II & Time & Filter I & Filter II \\ \hline 1 & 24 & 26 & 11 & 11 & 9 \\ 2 & 31 & 30 & 12 & 8 & 10 \\ 3 & 35 & 33 & 13 & 14 & 17 \\ 4 & 32 & 28 & 14 & 17 & 16 \\ 5 & 25 & 23 & 15 & 19 & 16 \\ 6 & 25 & 28 & 16 & 19 & 18 \\ 7 & 29 & 24 & 17 & 25 & 27 \\ 8 & 30 & 33 & 18 & 20 & 22 \\ 9 & 26 & 22 & 19 & 23 & 27 \\ 10 & 18 & 18 & 20 & 32 & 31 \\ \hline \end{tabular} a. Make a \(95 \%\) confidence interval for the mean of the population paired differences, where a paired difference is equal to the pollutant concentration passing through Filter I minus the pollutant concentration passing through Filter II. b. Using a \(5 \%\) significance level, can you conclude that the average paired difference for concentration levels is different from zero??

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