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Deceptive News. In the article "When News Reporters Deceive: The Production of Stereotypes" (Journalism & Mass Communication Quarterly, Vol. 84, No. 2, pp. 281-298), researchers D. Lasorsa and J. Dai investigated the relationship between authenticity and tone for news stories. Tone was measured by using a method of coding sentences and was categorized as positive, neutral, or negative. A sample of stories yielded the following data. At the 5% significance level, do the data provide sufficient evidence to conclude that an association exists between authenticity and tone?

Short Answer

Expert verified

The data provide sufficient evidence to conclude that an association exists between authenticity and tone at the 5 % significance level.

Step by step solution

01

Step 1. Given information

02

Step 2. Check whether or not the data provide sufficient evidence to conclude that an association exists between authenticity and tone.

Check whether or not the data provide sufficient evidence to conclude that an association exists between authenticity and tone.

Step 1:

The test hypotheses are given below:

Null hypothesis:

H0: There is no association exists between authenticity and tone.

Alternative hypothesis:

H1: There is an association exists between authenticity and tone.

Step 2:

Decide the level of significance.

Hence, the level of significance is α=0.05

Step 3:

Find the expected frequency and test statistic.

MINITAB procedure:

Step a: Choose Stat > Tables > Chi-Square test for association.

Step b: In Columns containing the table, enter the column of Authentic and Deceptive.

Step c: In Rows, select Tone.

Step d: Under Statistics, select Chi-square test, Display counts in each cell, Display marginal counts and expected cell counts.

Step e: Click OK.

Now,

Minitab output:

From the MINITAB output, the value of the chi-square statistic is 10.427.

03

Step 3. Continue to step 4 and 5

Step 4

Find the P-value.

From the MINITAB output, the P-value is 0.013.

Step 5:

Rejection rule:

If P-value≤a, then reject the null hypothesis.

Here, the P-value is lesser than the level of significance.

That is, P-value(=0.005)<α(=0.05).

Therefore, the null hypothesis is rejected at 5% level.

Thus, the results are statistically significant at 5%5% level of significance.

Thus, the data provide sufficient evidence to conclude that an association exists between authenticity and tone at the 5 % significance level.

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Most popular questions from this chapter

In each of the given Exercises, we have presented a contingency table that gives a cross-classification of a random sample of values for two variables, x, and y, of a population. For each exercise, perform the following tasks.

a. Find the expected frequencies. Note: You will first need to compute the row totals, column totals, and grand total.

b. Determine the value of the chi-square statistic.

c. Decide at the 5% significance level whether the data provide sufficient evidence to conclude that the two variables are associated.

US. Hospitals. Refer to Exercise 12.50.


24 or fewer25-7475 or moreTotal
General260158635575403
Psychiatric24242471737
Chronic132226
Tuberculosis0224
Other25177208410
Total310201042606580

a. Determine the conditional distribution of number of beds within each facility type.

b. Does an association exist between facility type and number of beds for U.S. hospitals? Explain your answer.

c. Determine the marginal distribution of number of beds for U.S. hospitals.

d. Construct a segmented bar graph for the conditional distributions and marginal distribution of number of beds. Interpret the graph in light of your answer to part (b),

e. Without doing any further calculations, respond true or false to the following statement and explain your answer. "The conditional distributions of facility type within number-of-beds categories are identical.

f. Determine the marginal distribution of facility type and the conditional distributions of facility type within number-of-beds categories.

g. What percentage of hospitals are general facilities?

h. What percentage of hospitals that have at least 75 beds are general facilities?

i. What percentage of general facilities have at least 75 beds?

Consider two χ2curves with degrees of freedom 12and 20respectively. which one more closely resembles a normal curve? Explain your answer.

Variegated Plants. Arabidopsis is a genus of flowering plants related to cabbage. A variegated mutant of the Arabidopsis has yellow streaks or marks. E. Miura et al. studied the origin of this variegated mutant in the article "The Balance between Protein Synthesis and Degradation in Chloroplasts Determines Leaf Variegation in Arabidopsis yellow variegated Mutants" (The Plant Cell, Vol. 19, No. 4, pp. 1313-1328). In a second-generation cross of variegated plants, 216 were variegated and 84 were normal. Genetics predicts that 75% of crossed variegated plants would be variegated and 25% would be normal. At the 10% significance level, do the data provide sufficient evidence to conclude that the second generation of crossed variegated plants does not follow the genetic predictions?

To decide whether two variables of a population are associated, we usually need to resort to inferential methods such as the chi-square independence test. Why?

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