/*! 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 227 Quiz Timing A young statistics p... [FREE SOLUTION] | 91Ó°ÊÓ

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Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

Short Answer

Expert verified
From the provided data, there is no sufficient evidence to support that the timing of the quiz (whether 'Early' or 'Late') significantly affects the mean grades.

Step by step solution

01

Analyze the data

The information provided from the Two-Sample T-Test is as follows: For the 'Late' class, we have a mean of 22.56 with a standard deviation of 5.13, while the 'Early' group had a mean of 19.73 with a standard deviation of 6.61. The difference between the 'Late' and 'Early' group means is estimated at 2.83. The tabular data also shows a 95% confidence interval for this difference, ranging from -0.20 to 5.86.
02

State the hypothesis

The null hypothesis (\(H_0\)) postulates that there is no significant difference between the mean grades of the 'Early' and 'Late' groups, i.e., the difference in their mean grades equals to 0. The alternative hypothesis (\(H_A\)) states that there is a difference, meaning that the difference in their mean grades does not equal to 0.
03

Consider the p-value

The p-value given in the exercise is 0.066. This value is compared against a significance level, usually set at 0.05. If the p-value is less than or equal to this level, then we reject the null hypothesis.
04

Conclusion

Since the p-value (0.066) is greater than the significance level (0.05), we fail to reject the null hypothesis. Therefore, based on this analysis, we don't have enough evidence to suggest that the mean grade depends on the timing of the quiz.

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Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is an essential method in statistics used to determine if there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. When a statistics professor wondered whether the timing of quizzes affected the performance of his students, he set up an experiment to test this.

In the example provided, hypothesis testing begins by establishing two opposing hypotheses. The null hypothesis, denoted as \(H_0\), represents the prediction that there is no effect or difference. It is essentially the status quo and is presumed true until evidence suggests otherwise. In our scenario, the null hypothesis claims that the quiz timing does not affect the students' mean grades.

The alternative hypothesis, \(H_A\), is what the researcher really wants to test; it indicates the presence of an effect or difference. Here, the alternative is that quiz timing does affect mean grades. The method used to test these hypotheses is a two-sample T-Test, which compares the means of two independent groups to see if there is statistically significant evidence that the associated population means are significantly different.

The next step is to compute a test statistic, which is a value calculated from the data that can be compared against a distribution to determine the likelihood of observing the data if the null hypothesis is true. The T-Value obtained from the computation in the solution is 1.87. This T-Value helps us understand where our observed difference lies within the context of the T-distribution.
P-Value Interpretation
The p-value is a crucial component in hypothesis testing and assists in deciding whether to reject the null hypothesis. It provides a measure of the strength of the evidence against the null hypothesis, indicating the probability of observing the data, or something more extreme, if the null hypothesis is indeed true.

In the educational exercise we're discussing, the p-value obtained is 0.066. Interpretation of the p-value depends heavily on the significance level (often denoted as \(\text{alpha}\)) chosen by the researcher, which is a threshold of probability used to determine whether the null hypothesis can be rejected. A commonly used significance level is 0.05, meaning that there is a 5% probability of rejecting the null hypothesis when it is actually true (a Type I error).

Since the p-value in this case is greater than 0.05, it suggests that the evidence against the null hypothesis isn't strong enough to reject it at the conventional 5% significance level. Nevertheless, the p-value isn't much larger; this may encourage the researcher to investigate further, possibly with a larger sample size or by considering other factors that could influence the outcomes.
Confidence Interval Analysis
A confidence interval provides a range of values, derived from the sample data, that is believed to cover the true population parameter with a specified level of confidence, typically 95%. In the context of the two-sample T-Test, a 95% confidence interval for the difference between means offers a range in which we expect the true difference between the population means to lie if the experiment were repeated multiple times.

The provided confidence interval in our example is (-0.20, 5.86). This states that we can be 95% confident that the true mean difference between quiz scores of the 'Late' and 'Early' groups lies within this range. Notably, the interval includes zero, which signifies that there is a possibility that no difference exists between the groups – supporting our previous failure to reject the null hypothesis from the p-value analysis.

Moreover, the confidence interval gives us more context than the simple binary outcome of the hypothesis test. The range tells us that if there is indeed an effect, it is not likely to be a large one, as the lower bound is very close to zero. When a confidence interval includes zero (in the context of mean differences), it emphasizes the potential for no effect size, which is an important insight for researchers and those applying statistics to real-world scenarios.

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Most popular questions from this chapter

When we want \(95 \%\) confidence and use the conservative estimate of \(p=0.5,\) we can use the simple formula \(n=1 /(M E)^{2}\) to estimate the sample size needed for a given margin of error ME. In Exercises 6.40 to 6.43, use this formula to determine the sample size needed for the given margin of error. A margin of error of 0.04 .

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To measure the effectiveness of a new teaching method for math in elementary school, each student in a class getting the new instructional method is matched with a student in a separate class on \(\mathrm{IQ}\), family income, math ability level the previous year, reading level, and all demographic characteristics. At the end of the year, math ability levels are measured again.

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

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