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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.

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.

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

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.

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.