Question

Denomination Effect Construct the confidence interval that could be used to test the claim in Exercise 1. What feature of the confidence interval leads to the same conclusion from Exercise 1?

Short answer

The confidence interval to test the given claim is equal to (–0.559, –0.175).

The presence of 0 in the confidence interval is the feature that is used to reach the conclusions.

Since the confidence interval of (–0.559, –0.175) does not contain the value 0, there is enough evidence to support the claim that money in a large denomination is less likely to be spent relative to the same amount of money in smaller denominations.

Step-by-step solution

Given information

The number of people who were given a $1 bill/4 quarters and who spent the money is tabulated along with the individual sample sizes.

Confidence interval

It is claimed that money in a large denomination is less likely to be spent relative to the same amount of money in smaller denominations.

That is, the proportion of people who receive a $1 bill and spend it is less than the proportion of people who receive 4quarters and spend the money.

The following formula is used to construct the confidence interval for estimating the difference in two population proportions:

\(CI = \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\)

Where,

• \({\hat p_1}\)denotes the sampleproportion of people who received a $1 bill and spent it.
• \({\hat p_2}\)denotes the sampleproportion of people who received 4 quarters and spent the money.
• E is the margin of error.

Calculation

The margin of error has the following formula:

\(E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \)

The following values are obtained from the previous exercise:

\({\hat p_1} = 0.261\)

\({\hat p_2} = 0.628\)

\(\begin{array}{l}{n_1} = 46\\{n_2} = 43\end{array}\)

The level of significance is equal to 0.05. Thus, the corresponding confidence level is equal to 95%.

The value of\({z_{\frac{\alpha }{2}}}\)when\(\alpha = 0.05\)is equal to 1.96.

Thus, the margin of error is equal to:

\(\begin{aligned}{c}E &= {z_{\frac{\alpha }{2}}}\sqrt {\frac{{{{\hat p}_1}{{\hat q}_1}}}{{{n_1}}} + \frac{{{{\hat p}_2}{{\hat q}_2}}}{{{n_2}}}} \\ &= \left( {1.96} \right)\sqrt {\frac{{\left( {0.261} \right)\left( {1 - 0.261} \right)}}{{46}} + \frac{{\left( {0.628} \right)\left( {1 - 0.628} \right)}}{{43}}} \\ &= 0.192\end{aligned}\)

Thus, the 95% confidence interval is equal to:

\(\begin{aligned}{c}CI &= \left( {\left( {{{\hat p}_1} - {{\hat p}_2}} \right) - E,\left( {{{\hat p}_1} - {{\hat p}_2}} \right) + E} \right)\\ &= \left( {\left( {0.261 - 0.628} \right) - 0.192,\left( {0.261 - 0.628} \right) + 0.192} \right)\\ &= \left( { - 0.559, - 0.175} \right)\end{aligned}\)

Therefore, the confidence interval to estimate the difference in the two proportions of people is equal to (–0.559, –0.175).

Interpretation of confidence interval

If the constructed confidence interval contains the value 0, there is not enough evidence to support the given claim.

If the constructed confidence interval does not contain the value 0, there is enough evidence to support the given claim.

Thus, the property of whether the confidence interval contains 0 is used to reach the conclusions.

Here, the confidence interval of (–0.559, –0.175) does not contain the value 0.

So, there is enough evidence to support the claim thatmoney in a large denomination is less likely to be spent relative to the same amount of money in smaller denominations.