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Reading In a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from th...

Question

Reading In a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from the Community College Survey of Student Engagement).(a) Determine a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.Solution:(b) Determine a 95% confidence interval for the proportion of community college students who have read a book for persona

Reading In a survey of 700 community college students, 481 indicated that they have read a book for personal enjoyment during the school year (based on data from the Community College Survey of Student Engagement).(a) Determine a 90% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.Solution:(b) Determine a 95% confidence interval for the proportion of community college students who have read a book for personal enjoyment during the school year.Solution:(c) What is the impact of increasing the level of confidence on the margin of error?Solution:



Answers


A Gallup poll conducted July 21-August $14,1978,$ asked 1006 Americans, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that $\bar{x}=18.8$ books and $s=19.8$ books. (a) Construct a $99 \%$ confidence interval for the mean number of books Americans read either all or part of during the preceding year. Interpret the interval. (b) Compare these results to those of Problem $13 .$ Were Americans reading more in 1978 than in $2005 ?$

The following is a solution to number 13 were given that the mean number of Americans in 2005 that red um either part or all of the book during the preceding year. So I guess 2004, um the sample there, so 1000 and six americans were sampled and that sample man was 13.4 in that standard deviation was 16.6, and we're asked to find the 99% confidence and over that. So we're gonna use the tea interval here because we don't know what the population standard deviation is, were given the sample standard deviation of 16.6, but we don't know the population standard deviation. So since we only know the s then we're gonna use that T interval and I'm gonna use technology here, but you can certainly use the formula or any other sort of technology one. I'm gonna use the T. I 84 because it works quite well. So if you go to stat and tests and we're gonna go down to this eighth option here, the T interval, so I'm gonna click eight And make sure the summary stats is highlighted, not data, because we're not giving any data here and I have everything already done here. So 13.4 is the X bar, 16.6 is the standard deviation and the sample size was 1006 and then that confidence level is .99 for 99%. And we calculate in this top band here 12.049 and 14.751, that's our confidence intervals. So let's go and write that down and then we interpret it. So it's between 12.049 and 14 point 751 books. So, the interpretation I'm actually gonna type out because it will be a little bit quicker. Here's generally how you were these interpretations, We can be Whatever.% 99 99% confident that the that true mean or the mean number of books americans read either all or part of during the preceding year Is between 12049 And 1475 1 books.

Well the data with N is equal to 1000 and six sample mean ex far equals 13.4 and standard deviation as for the sample, 16.6. We want to use this day to construct a 99% confidence interval for the population menu and interpret the results. So to start off with, because we have sample standard deviation, s not population standard deviation and we have angrier than equal to 30. We can use a student's t distribution to solve this problem. That means that we need to identify the T score hasn't written here for the degree of freedom of approximately 1000 with a probability T falls between the negative and positive critical value. T C is 0.99 So this corresponds using a tea table which maps for particular degrees of freedom, the probability to the T score T C equals 2.581 Now we can use this to construct our margin of error equals T c 10 s over root end plugging in our T c s N gives us equals 1.35 And we construct the confidence interval. Use the following formula where x y minus E is less than you earn less than X Y plus E, giving us our confidence interval of 12.5 is less than you as less than 14.75 Finally, we can interpret these findings to suggest that We have with 99 confidence based on of our data set, that the mute or rather population mean falls between 12.05 and 14.75.


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