Question

This Questión pt My Courses Course Home The data given to the right includes data from 44 candies, and 10 of them are red. The company that makes the candy claims that 32 % of its candies are red. Use the sample data to construct a 90% confidence interval estimate of the percentage of red candies. What do you conclude about the claim of 32 % ? Weights (g) of a Sample Bag of Candy Announcements Yellow 0.764 0.935 Blue Brown Green Red 0.915 0,987 0259 0.993 0.702 0.923 0.939 0.859 0.792 0.824 0.727 0.844 0.808 0.897 Assignments 0813 0.761 0.727 0.739 0.939 0.795 0.904 Construct a 90 % confidence interval estimate of the population percentage of candies that are red 0.737 0.885 0.745 0783 Study Plan 0.887 0.902 0.924 0.793 0.772 0.715 0.879 0.735 0.805 '%<p< (Type an integer or decimal rounded to one decimal place as needed.) 0.753 0.721 0.995 0716 0.863 0.832 Gradebook 0.882 0.993 Is the result consistent with the 32% rate that is reported by the candy maker? StatCrunch O No, because the confidence interval does not include 32 % . еТеxt Yes, because the confidence interval indudes 32% . Chapter Contents Tools for Success Click to select your answerls). Multimedia Librar Sampie tess ana quizzes can be taken Tor practcr ar to pura yyour stuay pran. sampie tests and quiz2es donot aTrecr your grare Purchase Options O Type here to search STE seT F7 FR FB % LOC 7 7 P п _un

Answer 1

Solution :

Given that,

x - 10

n - 44

Point estimate ($\hat p$) = x / n = 10/44 = 0.23

1 - $\hat p$ = 1 - 0.23 = 0.77

Z_$\alpha$/2 = 1.645

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$[$\hat p$ * (1 - $\hat p$) / n]

= 1.645 * $\sqrt$[(0.23 * 0.77) / 44]

= 0.10

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.23 - 0.10 < p < 0.23 + 0.10

0.13 < p < 0.33

The 90% confidence interval for the population proportion p is from 13 % to 33%

Yes. The result consistent with the 32% rate because it includes in confidence interval.