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Chapter 10: Problem 30

10.30 Does a content title help a reader comprehend a piece of writing? Twenty-six participants were given an article to read without a title. They then rated themselves one their comprehension of the information on a scale from 1 to \(10,\) where 10 was complete comprehension. The same 26 participants were then given the article again, this time withe an appropriate title, and asked to rate their comprehension The resulting summarized data were given as \(\bar{d}=4.76\) and \(s_{d}=2.33,\) where \(d=\) rating with title - rating without title Comprehension was generally higher on the second reading than on the first by an average of 3.2 on this scale. Does this sample provide sufficient evidence that a content title does make a difference with respect to comprehension? Use \(\alpha=0.05\)

Short Answer

Expert verified
We need to calculate the t-value and compare it to the critical t-value. If the calculated t-value is greater than the critical t-value, we reject the null hypothesis, which means the content title does make a significant difference with respect to comprehension. If the t-value is less than the critical t-value, we fail to reject the null hypothesis, meaning the content title does not make a significant difference with respect to comprehension.

Step by step solution

01

Set up the Null and Alternative Hypothesis

The null hypothesis \(H_0\) asserts that the mean difference (d) between the two measurements is zero. This means, there is no significant difference in readers comprehension whether the article has a title or not: \(H_0: \mu_d = 0\).\n The alternative hypothesis \(H_1\) asserts that there is a significant difference: \(H_1: \mu_d \neq 0\). \nThe level of significance or alpha (\(\alpha\)) is given as 0.05.
02

Calculate the t-value

We are given that \(\bar{d}=4.76\) is the sample mean difference and \(s_{d}=2.33\) is the standard deviation of the differences. The sample size n is 26 (the number of participants). The t-value can be calculated as follows: \(t = \frac{\bar{d} - \mu_d}{s_{d}/\sqrt{n}} = \frac{4.76 - 0}{2.33/\sqrt{26}}\).
03

Determine the Critical t-value

For a two-tailed test with 25 degrees of freedom (26 - 1) and an alpha level of 0.05, we find the critical t-value from a t-distribution table or using a calculator. The critical t-value for a two-tailed test with df = 25 and \(\alpha=0.05\) is approximately 卤2.060.
04

Make a Decision

If the calculated t-value is greater than the critical t-value, then we reject the null hypothesis. If the calculated t-value is less than the critical t-value, then we fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
A null hypothesis is a statement made during hypothesis testing. It suggests that there is no effect or no difference in a given situation. For this exercise, the question was whether adding a content title to an article changes how well people understand what they are reading. The null hypothesis, symbolized as \(H_0\), claims that adding a title makes no difference to readers' comprehension. Mathematically, it states that the mean difference \(\mu_d\) between comprehension with and without the title is zero: \(H_0: \mu_d = 0\).
The null hypothesis serves as a baseline to challenge with evidence. We test it to see if a significant difference is present when a title is added. If evidence supports a difference, we may reject the null hypothesis. If the evidence does not support a difference, we do not reject the null hypothesis, meaning we accept that the title has no impact on comprehension.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis. It represents what we expect to prove with statistical evidence. In this exercise, the alternative hypothesis is symbolized by \(H_1\). It suggests that there is a significant difference in comprehension scores when a title is present versus when it is absent.
The alternative hypothesis for this test is written as \(H_1: \mu_d eq 0\). This means that the mean difference in comprehension is not zero. In simpler terms, the comprehension rate changes when a title is added. The goal of hypothesis testing is ultimately to see if the data collected provides enough evidence to support this alternative hypothesis. For this test, an alpha level of 0.05 is set as the threshold for determining statistical significance.
T-value Calculation
Calculating the t-value is a crucial part of hypothesis testing. It is used to determine the relationship between the sample data and the null hypothesis. Here, we want to find if the difference in comprehension with and without a title is significant.
To calculate the t-value, we use the formula: \[ t = \frac{\bar{d} - \mu_d}{s_{d}/\sqrt{n}} \] where:
  • \(\bar{d}\) is the sample mean difference, which is 4.76 in this case.
  • \(\mu_d\) is the population mean difference under the null hypothesis, assumed to be 0.
  • \(s_{d}\) is the standard deviation of the differences, equal to 2.33.
  • \(n\) is the sample size, which is 26 (the number of participants).
The calculated t-value helps us decide if the null hypothesis should be rejected. If the t-value falls outside the range determined by the critical t-value (here, approximately 卤2.060 for 25 degrees of freedom), it indicates the sample provides sufficient evidence to support the alternative hypothesis.

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Most popular questions from this chapter

Ten recently diagnosed diabetics were tested to determine whether an educational program would be effective in increasing their knowledge of diabetes. They were given a test, before and after the educational program, concerning self-care aspects of diabetes. The scores on the test were as follows: $$\begin{array}{lcccccccccc}\text { Patient } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \text { Before } & 75 & 62 & 67 & 70 & 55 & 59 & 60 & 64 & 72 & 59 \\\\\text { Afler } & 77 & 65 & 68 & 72 & 62 & 61 & 60 & 67 & 75 & 68 \\\\\hline\end{array}$$The following MINITAB output may be used to determine whether the scores improved as a result of the program. Verify the values shown on the output [mean difference(MEAN), standard deviation of the difference (STDEV), standard error of the difference (SE MEAN), \(t \star\) (T-Value), and \(p\) -value] by calculating the values yourself.

Are females as serious about golf as men are? If so, would the price of a driver for a man be the same as the price of a driver for a female? It was contended that women's drivers would be cheaper. Random samples of drivers were taken from the Golflink.com website. The prices were: $$\begin{array}{llllll} \text { Male } & & & & & \\\\\hline 149.99 & 299.99 & 49.99 & 499.99 & 167.97 & 299.99 \\\399.99 & 199.99 & 99.99 & 149.99 & & \\\\\hline \text { Female } & & & & & \\\\\hline 199.99 & 79.99 & 499.99 & 199.97 & 299.99 & 99.99\end{array}$$ At the 0.05 level of significance, is there sufficient evidence to support the contention that men's drivers are more expensive than women's drivers? Assume normality of golf driver prices.

10.86 Calculate the maximum error of estimate for a \(90 \%\) confidence interval for the difference between two proportions for the following cases:a. \(\ n_{1}=40, p_{1}^{\prime}=0.7, n_{2}=44,\) and \(p_{2}^{\prime}=0.75\) b. \(\ n_{1}=36, p_{1}^{\prime}=0.33, n_{2}=38,\) and \(p_{2}^{\prime}=0.42\)

When a hypothesis test is two-tailed and Excel is used to calculate the \(p\) -value, what additional step must be taken?

State the null hypothesis, \(H_{o},\) and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. There is an increase in the mean difference between post-test and pre-test scores. b. Following a special training session, it is believed that the mean of the difference in performance scores will not be zero. c. On average, there is no difference between the readings from two inspectors on each of the selected parts. d. The mean of the differences between pre-self-esteem and post-self-esteem scores showed improvement after involvement in a college learning community.
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