Question
A research center conducted a national survey about teenage behavior. Teens were asked whether they had consumed a soft drink in the past week. The following table shows the counts for three independent random samples from major cities.
(a) Suppose one teen is randomly selected from each city’s sample. A researcher claims that the likelihood of selecting a teen from Baltimore who consumed a soft drink in the past week is less than the likelihood of selecting a teen from either one of the other cities who consumed a soft drink in the past week because Baltimore has the least number of teens who consumed a soft drink. Is the researcher’s claim correct? Explain your answer.
(b) Consider the values in the table. (i) Construct a segmented bar chart of relative frequencies based on the information in the table.
(ii) Which city had the smallest proportion of teens who consumed a soft drink in the previous week? Determine the value of the proportion.
(c) Consider the inference procedure that is appropriate for investigating whether there is a difference among the three cities in the proportion of all teens who consumed a soft drink in the past week.
(i) Identify the appropriate inference procedure.
(ii) Identify the hypotheses of the test.
▶️Answer/Explanation
Ans:
The researcher’s claim is incorrect. Just because the count of Baltimore teens who consumed a soft drink in the past meek doesn’t mean it is leas likely to select a teen from Baltimore who answered yes than the other cities. In reality, the likelihood of a Baltimore teen answering yes is \(\frac{727}{904}=0.804\), which is higher than Detroit \(\left(\frac{1233}{1662}=0.742\right)\) and San Diego \(\left(\frac{1482}{2250}=0,650\right)\). There were differing numbers of teens surveyed in the three cities, so the numbers cannot be compared directly; theymust be proportions first.
(b)
San Diego \(\rightarrow p=\frac{1482}{2280}=0.65\)
(c) The appropriate inference procedure is a \(x^2\)-test for homogeneity.
\(H_s:\) All three cities have the same proportion of teens who consumed a soft drink in the past week.
\(H_a\) : The proportion of teens who had a soft drink in the past week differs among the three cities,
Question
The manager of a local fast-food restaurant is concerned about customers who ask for a water cup when placing an order but fill the cup with a soft drink from the beverage fountain instead of filling the cup with water. The manager selected a random sample of 80 customers who asked for a water cup when placing an order and found that 23 of those customers filled the cup with a soft drink from the beverage fountain.
(a) Construct and interpret a 95 percent confidence interval for the proportion of all customers who, having asked for a water cup when placing an order, will fill the cup with a soft drink from the beverage fountain.
(b) The manager estimates that each customer who asks for a water cup but fills it with a soft drink costs the restaurant \(\$ 0.25\). Suppose that in the month of June 3,000 customers ask for a water cup when placing an order. Use the confidence interval constructed in part (a) to give an interval estimate for the cost to the restaurant for the month of June from the customers who ask for a water cup but fill the cup with a soft drink.
▶️Answer/Explanation
Ans:
(a)
Step li eel \(p=\) the proportion of all customers who fill the cup with soda rather than water when asking for water
Step 2: 1 proportion 2 interval
1. The manager selected a random sample \(\checkmark\)
2. independent: ion \(<\) pop \(10(80)=800 \quad 800<p o p \checkmark\)
3. Normal: \(n p>10 \quad n(1-p)>10\)
$
(.2875)(80)=23 \quad a(1-.2875)(80)=57
$
Step 3: \((.18832, .38668)\)
Step 4: we are \(95 \%\) confident that the interval from .88332 to .38668 will capture the true population proportion of all customers who fill the cup with soda rather than water when asking for water.
(b)$
\begin{array}{ll}
3.000(.18832)=564.96 & 564.96(.25)=141.24 \\
3.000(.38668)=1160.04 & 1160.04(.25)=290.01
\end{array}
$
in the month of June, the cost to the restaurant will be between \(\$ 141.24\) and \(\$ 290.0\) is using the \(95 \%\) confidence interval from part (a).