Nominal Data. In Exercises 9–12, use the sign test for the claim involving nominal data.

Stem Cell Survey Newsweek conducted a poll in which respondents were asked if they “favor or oppose using federal tax dollars to fund medical research using stem cells obtained from human embryos.” Of those polled, 481 were in favor, 401 were opposed, and 120 were unsure. Use a 0.01 significance level to test the claim that there is no difference between the proportions of those opposed and those in favor.

The number of respondents who favor and oppose a particular policy is given to be equal to 481 and 401, respectively.

The researcher wants to test the claim that there is no difference between the proportions of those opposed and those in favor of using the federal tax dollars to fund medical research.

The significance level is 0.01.

The sign test is known as the non-parametric test that can be used when the data is nominal, involving only two categories. In this test, one category is represented by a positive sign, and a negative sign represents other category.

If the sample size (n) is at most 25, then the test statistic is represented by “x” which means thenumber of times the less frequent sign (positive or negative) occurs.

If the sample size (n) is more than 25, then the test statistic is given as:

\(z = \frac{{\left( {x + 0.5} \right) - \frac{n}{2}}}{{\frac{{\sqrt n }}{2}}}\)

Considering that the proportion of those who oppose and those who are in favor should be the same, the null hypothesis is as follows:

There is no difference between the proportions of those opposed and those in favor

The alternative hypothesis is as follows:

There is a difference between the proportions of those opposed and those in favor.

A negative sign denotes the respondents who oppose.

A positive sign denotes the respondents who are in favor.

The number of positive signs = 481.

The number of negative signs = 401.

Since it can be seen that the number of positive signs and negative signs is not the same, the observation does not contradict the alternative hypothesis.

The sample size (n) is equal to:

\(\begin{array}{c}n = 481 + 401\\ = 882\end{array}\)

Let x be the number of times the less frequent sign occurs.

The less frequent sign is the negative sign corresponding to the number of people who opposed the policy.

The value of x is equal to 401.

As the sample size n is greater than 25, the value of z is calculated.

The test statistic z is calculated as shown:

\(\begin{array}{c}z = \frac{{\left( {x + 0.5} \right) - \frac{n}{2}}}{{\frac{{\sqrt n }}{2}}}\\ = \frac{{\left( {401 + 0.5} \right) - \frac{{882}}{2}}}{{\frac{{\sqrt {882} }}{2}}}\\ = - 2.66\end{array}\)

P-value:

Referring to Table A-2, the p-value obtained by using the test statistic is 0.0078.

Since the p-value is less than the significance level, so the decision is to reject the null hypothesis.

Critical value:

The critical value of z from the standard normal table for a two-tailed test with a value of\(\alpha \)= 0.01 is equal to\( \pm 2.575\).

Since the absolute value of zequal to 2.66 is greater than the critical value, the null hypothesis is rejected.

There is enough evidence to conclude that there is a difference in the proportion of respondents in favour and those opposed.