Question

Consider the following two questions designed to assess quantitative literacy:

a. What is \[15\% \]of 1000?

b. A store is offering an \[15\% \]off sale on all TVs. The most popular television is normally priced at $1000. How much money would a customer save on the television during this sale?

Suppose the first question is asked of 200 randomly selected college students, with 164 answering correctly; the second one is asked of a different random sample of 200 college students, resulting in 140 correct responses (the sample percentages agree with those given in the article "Using the Right Yardstick: Assessing Financial Literacy Measures by Way of Financial Well-Being," J. of Consumer Affairs, 2013: 243-262; the investigators found that those who answered such questions correctly, particularly questions with context, were significantly more successful in their investment decisions than those who did not answer correctly). Carry out a test of hypotheses at significance level 0.05 to decide if the true proportion of correct responses to the question without context exceeds that for the one with context.

Short answer

150

There is sufficient evidence to support the claim that the true proportion of correct responses to the question without context exceeds that for the one with context.

Step-by-step solution

To find what is 15% of 1000

Given:

15% or 1000

Rewrite the given expression as a product:

\[15\% \times 1000\]

Rewrite the percentage as a decimal, by removing the percentage sign and dividing the numeral by 100:

\(\frac{{15}}{{100}} \times 1000\)

Divide 1000 by 100:

\(15 \times 10\)

Evaluate the product:

150

Thus 15% of 1000 is 150.

Note: the hypothesis test mentioned after part (b) has been in part (b) of this exercise.

To find how much money would a customer save on the television during this sale

The amount saved is the product of the percentage and the price.

\(15\% \times 1000\)

Rewrite the percentage as a decimal, by removing the percentage sign and dividing the numeral by 100:

\(\frac{{15}}{{100}} \times 1000\)

Divide 1000 by 100:

\(15 \times 10\)

Evaluate the product:

150

Thus 15% of 1000 is 150.

Given:

\(\begin{array}{l}{x_1} = 164\\{n_1} = 200\\{x_2} = 140\\{n_2} = 200\\\alpha = 0.05\end{array}\)

Given claim: Exceeds

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\{H_a}:{p_1} > {p_2}\end{array}\)

To determine the value of the test statistics

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}{{\hat p}_1} = \frac{{{x_1}}}{{{n_1}}} = \frac{{164}}{{200}} \approx 0.82\\{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}} = \frac{{140}}{{200}} \approx 0.7\\{{\hat p}_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}} = \frac{{164 + 140}}{{200 + 200}} = \frac{{304}}{{400}} \approx 0.76\end{array}\)

Determine the value of the test statistic:

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }} = \frac{{0.82 - 0.7}}{{\sqrt {0.76(1 - 0.76)} \sqrt {\frac{1}{{200}} + \frac{1}{{200}}} }} \approx 2.81\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. Determine the P-value using table the normal probability table in the appendix:

\(P = P(Z > 2.81) = 1 - P(Z < 2.81) = 1 - 0.9975 = 0.0025\)

If the P-value is smaller than the significance level, then reject the null hypothesis:

\(P < 0.05 \Rightarrow {\rm{ Reject }}{H_0}\)

There is sufficient evidence to support the claim that the true proportion of correct responses to the question without context exceeds that for the one with context.

Final proof

150

There is sufficient evidence to support the claim that the true proportion of correct responses to the question without context exceeds that for the one with context.