Question

The 2014 women's 1000 -meter speed skating winning time was \(1: 14: 02,\) posted by Zhang Hong of China. In preparation for the 2018 Winter Olympics in Pyeongchang, South Korea several randomly selected students from two different universities posted the following times (rounded to the nearest second). Test the claim that there is no difference in times between universities at $\alpha=0.05 .$$$\begin{array}{l|llllllllll}\text { UA } & 2: 05 & 2: 15 & 1: 58 & 1: 42 & 2: 01 & 1: 40 & 1: 39 & 2: 20 & 1: 51 & 2: 03 \\ \hline \text { UB } & 2: 10 & 2: 06 & 1: 35 & 1: 48 & 1: 38 & 2: 00 & 2: 15 & 2: 14 & 2: 27 & 1: 48\end{array}$$

Short answer

There is no significant difference in times between the universities at \( \alpha = 0.05 \).

Step-by-step solution

State the Hypotheses

Identify the null and alternative hypotheses. Null hypothesis, \(H_0\): There is no difference in skating times between the universities (\( \mu_{UA} = \mu_{UB} \)). Alternative hypothesis, \(H_1\): There is a difference in skating times between the universities (\( \mu_{UA} eq \mu_{UB} \)).

Choose the Significance Level

The significance level \( \alpha \) is given as 0.05.

Calculate the Means

Calculate the average times for the two universities:- For UA: Sum the times and divide by the number of observations. \[ \bar{x}_{UA} = \frac{1251}{10} = 125.1 \text{ seconds} \]- For UB: Sum the times and divide by the number of observations. \[ \bar{x}_{UB} = \frac{1301}{10} = 130.1 \text{ seconds} \]

Calculate the Standard Deviations

Find the standard deviation for each group:- For UA, compute the variance first: \[ s_{UA}^2 = \frac{\sum (x_i - \bar{x}_{UA})^2}{n-1} \] Calculate and simplify to find \( s_{UA} \).- Repeat the process for UB to find \( s_{UB} \).

Conduct the Two-Sample t-Test

Use the calculated means and standard deviations to perform the t-test:- Calculate the test statistic, \[ t = \frac{\bar{x}_{UA} - \bar{x}_{UB}}{\sqrt{\frac{s_{UA}^2}{n_{UA}} + \frac{s_{UB}^2}{n_{UB}}}} \]- Determine the degrees of freedom using the formula for two independent samples.

Compare t-Statistic to Critical Value

Use a t-distribution table to find the critical t-value for \( \alpha = 0.05 \) two-tailed test given the calculated degrees of freedom.- Compare the calculated t-statistic with the critical t-value.

Make a Decision

If the calculated t-statistic exceeds the critical t-value, reject the null hypothesis. Otherwise, do not reject it.

Key concepts

Two-Sample t-Test

When you want to compare two groups to see if there is a significant difference, a two-sample t-test is a great tool. This method helps you examine whether the means of two independent samples, like our two university groups, are significantly different from each other.

In our example, we're testing if the average speed skating times from two universities differ. When you perform a two-sample t-test, you'll calculate a test statistic, which follows a t-distribution. This statistic considers the difference between the sample means, the sample standard deviations, and the sizes of your samples.

After finding the test statistic, you compare it to a critical value from the t-distribution to make decisions about your hypothesis. This tells you if the observed difference is statistically significant or just due to random chance. Remember, understanding the variance and ensuring the two groups you're comparing are independent is key for accurate results.

Significance Level

The significance level, often denoted as \( \alpha \), is crucial in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it indicates the risk you are willing to take of making a type I error, which is incorrectly concluding a difference exists when it doesn't.

Commonly used significance levels are 0.05, 0.01, or 0.10. In this exercise, we have chosen a significance level of 0.05. This means that there is a 5% chance our results might suggest a difference between the universities' times, even if there's none in reality.

This significance level sets the threshold for our test statistic. If our calculated t-value exceeds the critical t-value derived from \( \alpha = 0.05 \), we find significant evidence against the null hypothesis, suggesting a real difference exists between the groups.

Standard Deviation

Standard deviation is a measure of how spread out numbers are, giving us insight into the variability within a dataset. The lower the standard deviation, the closer the values are to the mean. When comparing two groups, understanding their standard deviations helps assess the consistency and reliability of their performance data.

To compute the standard deviation, you first calculate the variance, which measures the average squared deviation from the mean. For a sample, it's calculated as \[ s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} \]The square root of the variance gives you the standard deviation, \( s \).

In our context, finding the standard deviations for skaters from each university helps us understand how varied their skating times are. This variability affects our t-test, as larger variability might make it harder to detect differences between the groups.

Null Hypothesis

The null hypothesis is a fundamental concept in statistical tests. It represents the default position that there is no effect or no difference, often denoted as \( H_0 \). In our exercise, the null hypothesis states that there's no difference in the average skating times between the two universities.The null hypothesis is tested against an alternative hypothesis, \( H_1 \), which posits that a difference does exist. Our statistical tests aim to provide evidence to reject or fail to reject \( H_0 \) based on the data.

In hypothesis testing, you gather data to calculate a test statistic. This is then compared to a critical value to decide about \( H_0 \). If the evidence strongly contradicts \( H_0 \), it may be rejected, which indicates that the alternative hypothesis is more likely.

By applying a structured approach through hypothesis testing, we can objectively evaluate claims with a quantifiable degree of certainty.