Math  /  Data & Statistics

QuestionThere is a popular lottery in which a ticket is called a scratcher. An advertisement for this lottery claims that 55%55 \% of the population of all the scratchers are winning ones. You want to research this claim by selecting a random sample of 35 scratchers.
Follow the steps below to construct a 90%90 \% confidence interval for the population proportion of all winning scratchers. Then state whether the confidence interval you construct contradicts the advertisement's claim. (If necessary, consult a list\underline{l i s t} of formulas.) (a) Click on "Take Sample" to see the results from the random sample. \begin{tabular}{|c|c|c|c|} \cline { 3 - 4 } & & Number & Proportion \\ \hline Take Sample & Winning scratcher & 14 & 0.4 \\ \hline Losing scratcher & 21 & 0.6 \\ \hline \end{tabular}
Enter the values of the sample size, the point estimate of the population proportion, and the critical value you need for your 90%90 \% confidence interval. (Choose the correct critical value from the table of critical values provided.) When you are done, select "Compute".
Sample size: \square Standard error:
Point estimate: \square Margin of error: Critical values z0.005=2.576z_{0.005}=2.576 z0.010=2.326z_{0.010}=2.326 Critical value: z0.025=1.960z_{0.025}=1.960

Studdy Solution
Interpret the confidence interval in relation to the advertisement's claim:
The advertisement claims that 55% of the scratchers are winning ones. The confidence interval we calculated is (0.2676,0.5324) (0.2676, 0.5324) , which does not include 0.55. Therefore, the confidence interval contradicts the advertisement's claim.
The confidence interval is:
(0.2676,0.5324) (0.2676, 0.5324)
The confidence interval does not support the advertisement's claim.

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