91Ó°ÊÓ

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:

$H_0$: There is no association exists between authenticity and tone.

Alternative hypothesis:

$H_1$: There is an association exists between authenticity and tone.

Step 2:

Decide the level of significance.

Hence, the level of significance is $\mathrm{Î\pm }=0.05$

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.

Step 4

Find the P-value.

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

Step 5:

Rejection rule:

If $P\text{-±¹²¹±ô³Ü±ð≤a}$, then reject the null hypothesis.

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

That is, $P\text{-value}(=0.005)<\mathrm{Î\pm }(=0.05)$.

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.