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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.

Step by step solution

01

Calculate the Point Estimate of the Difference

The point estimate of the difference in population means is simply the difference in sample means. For the tenth-graders and the twelfth-graders, this is \(29 - 22 = 7\) hours.

02

Calculate the standard error

Standard error(SE) is calculated using the formula \(\sqrt{(s1^2/n1) + (s2^2/n2)}\), where \(s1 = 7.0\) and \(s2 = 6.2\) are the sample standard deviations, and \(n1 = 25\) and \(n2 = 23\) are the sample sizes. Substituting the given values, SE = \(\sqrt{(7.0^2/25) + (6.2^2/23)}\).

03

Calculate the Confidence Interval

The 90% confidence interval is given by \(Point Estimate \pm T_{critical} \times SE\). Given that the degree of freedom is \(n1+n2-2 = 46\), and we are looking for a 90% confidence interval, \(T_{critical}\) is approximately 1.68, you can find this value in t-distribution table. Substituting the calculated values, the confidence interval becomes \[(7 - 1.68 \times SE, 7 + 1.68 \times SE)\].

04

Hypothesis Testing

The null hypothesis is that the mean amounts of free time per week for tenth and twelfth graders is the same (\(μ1=μ2\)), and the alternative hypothesis is that the means are different (\(μ1≠μ2\)). We calculate the t-statistic as \(t = \frac{Point Estimate – Hypothesized Difference}{SE} = \frac{7 – 0}{SE}\). If this t-statistic is greater than \(T_{critical}\), we reject the null hypothesis. If it is less than \(T_{critical}\), we fail to reject the null hypothesis. Assuming unequal variances, the critical t-value with 54 degrees of freedom and 5% significance level is approximately \(\pm2.00\).

05

Conclusion

Compare the calculated t-statistic and the critical t-value. If the absolute value of our test statistic is greater than the critical t-value, reject the null hypothesis and conclude that there is sufficient evidence at the \(\alpha = 0.05\) level to conclude that there is a difference in mean free time for tenth and twelfth graders. If the absolute value of the test statistic is not larger than the critical t-value, fail to reject the null hypothesis and conclude that there is not sufficient evidence to conclude that there is a difference.

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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 make decisions about the characteristics of a population based on a sample. It's a critical concept in statistics when trying to determine if there's a statistically significant effect or difference.

Here's a simple breakdown of the process:

• Start with a null hypothesis (that there is no effect or difference) and an alternative hypothesis (that there is an effect or difference).
• Use statistical tests to determine if the observed data is consistent with the null hypothesis.
• Choose a significance level (like 5%, denoted as \(\alpha = 0.05\)) to determine how strong the evidence must be to reject the null hypothesis.
• Calculate a test statistic (like a t-statistic) to see how far the sample data diverges from the null hypothesis.
• Use this test statistic to conclude whether to reject or fail to reject the null hypothesis.

In the given exercise, the counselor is testing whether the mean free time for tenth-graders differs from that of twelfth-graders.

Sample Mean

The sample mean is a measure of the central tendency of a sample. It is the average of all data points collected in the sample.

To calculate the sample mean:

• Add all the observations together.
• Divide the sum by the number of observations.

In simple terms, it provides an estimate of what the average value might be if you measured every single person in the population. For instance, in the exercise:

• The mean free time for tenth-graders is 29 hours per week.
• The mean for twelfth-graders is 22 hours per week.

Thus, the point estimate of the difference in means between grades is given as 29 - 22 = 7 hours.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means the data points are close to the mean, while a high standard deviation means they are spread out over a wide range.

It's important to understand:

• How it helps in understanding the spread of the data in a sample.
• In the context of the exercise:
  • The standard deviation for tenth-graders is 7.0 hours.
  • The standard deviation for twelfth-graders is 6.2 hours.
• Standard deviation is used in calculating the standard error, which is crucial for constructing confidence intervals and hypothesis tests.

In this problem, standard deviation helps in assessing how consistent free time measurements are within each grade level.

T-distribution

The t-distribution is a probability distribution used when estimating population parameters when the sample size is small, and the population variance is unknown.

Key characteristics of the t-distribution include:

• It is similar to the normal distribution but has thicker tails, providing for more variability, which is important for small sample sizes.
• The degree of freedom, determined by the size of the sample, impacts the shape of the t-distribution.

When applying it in the exercise:

• The critical t-value was used to construct a confidence interval around the difference in sample means.
• It's also used in hypothesis testing to decide whether to reject the null hypothesis.
• With 46 degrees of freedom, a t-value of about 1.68 is found from the t-table for a 90% confidence interval.

In conclusion, understanding t-distribution helps in making reliable inferences from sample data about the population.