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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 the \(10 \%\) significance level, can you conclude that the average number of Christmas cards mailed out by all households in Town A is different from the corresponding average for Town B?

Step by step solution

01

Identify the samples and their statistics

We have two samples, one from Town A and another from Town B. Sample statistics for Town A are - sample size, n1 = 80, sample mean, \(\bar{x1} = 28.55\), and sample standard deviation, \(s1 = 10.30\). For Town B, the sample statistics are - sample size, n2 = 58, sample mean, \(\bar{x2} = 33.67\), and sample standard deviation, \(s2 = 8.97\).

02

Construct a 95% confidence interval for the difference in the means

We use the formula \(\bar{x1} - \bar{x2} \pm z\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\) for confidence interval estimation. Here, \(\bar{x1} = 28.55\), \(\bar{x2} = 33.67\), \(s1 = 10.30\), \(s2 = 8.97\), \(n1 = 80\), \(n2 = 58\) and z is the z-score for 95% confidence level which is 1.96. Plug the values and calculate the confidence interval.

03

Conducting Hypothesis testing

We will conduct a two-sample z test at the 10% significance level. First, state the null hypothesis H0 (the mean number of cards mailed in Town A and Town B is the same) and alternative hypothesis H1 (The average number of cards mailed is different in both towns). Then calculate the test statistic using the formula \(z = \frac{(\bar{x1} - \bar{x2}) - 0}{\sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}}\). The denominator is the pooled standard deviation. If the result from the test statistic calculation (z) is either less than the negative of the critical z value for 10% significance level (-z_alpha/2 = -1.645) or greater than the positive critical z value (z_alpha/2 = 1.645), then we will reject the null hypothesis. Else, we will not 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 Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions or inferences about a population based on sample data. It begins with setting up two opposing hypotheses. Here's a breakdown:

• **Null Hypothesis (H鈧�):** This statement assumes no effect or difference. In our exercise, it suggests that the average number of Christmas cards mailed in Town A is the same as Town B.
• **Alternative Hypothesis (H鈧�):** This proposes a difference. Here, it suggests that the average number is different between the two towns.

Next, we use sample data to calculate a test statistic, which helps us determine whether to reject the null hypothesis. Depending on the significance level, we decide if there's enough evidence to claim a difference.
It鈥檚 crucial to understand that rejecting the null doesn鈥檛 prove the alternative; it only suggests there's enough evidence against the null. This process helps researchers make informed decisions based on data.

Exploring the Two-Sample Z Test

The two-sample z test is used to compare the means from two different populations. It鈥檚 especially useful when the standard deviations are known or assumed to be equal.

Here鈥檚 how it works:

• **Sample Information:** Gather data from both samples, like means and standard deviations. In our case, Town A and Town B have different sample sizes, means, and standard deviations.
• **Calculate the Test Statistic:** The formula used is \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]This compares the difference in sample means to zero, adjusted for the pooled standard deviation.
• **Decision Rule:** Compare the calculated z value with critical values based on the significance level (e.g., 10%). If it falls beyond these critical values, the null hypothesis is rejected.

By following these steps, the two-sample z test helps determine if two population means are significantly different, guiding data-driven decisions.

What is Statistical Significance?

Statistical significance is a way to assess whether the results from an experiment or test are likely to be true, rather than occurring by random chance.

To determine statistical significance, we consider:

• **Significance Level (伪):** This is the threshold for rejecting the null hypothesis. Common levels are 0.05 or 0.10. For our test, a 10% level (\(\alpha = 0.10\)) was used.
• **p-Value:** This tells us the probability of obtaining the observed results if the null hypothesis is true. A smaller p-value indicates stronger evidence against the null.
• **Critical Values:** Based on significance level, critical values help determine the rejection region for the null hypothesis.

The aim is to determine if observed differences (like mailing trends in Town A and B) are meaningful or just due to random variations. Statistical significance helps clarify if findings are real and worth further investigation.