Question

1. A sample of 25 adults living in Kentucky had an average IQ of 102. The margin of error for a 95% confidence interval is 18. We are interested in a 99% confidence interval for the average IQ of all adults in Kentucky.

2. We are interested in a 90% confidence interval for the proportion of EKU students who will spend summer break outside of Kentucky if the sample had 75 students of which 20 would be leaving the state. The margin of error for a 90% confidence interval was found to be .084. (Round your answer to 3 decimal places.

3. Suppose that a student is working on a statistics project and has collected data on pulse rates from a random sample of 30 students from her college. From her sample she finds an average of 68.65 beats per minute. The margin of error for a 95% confidence interval is 3.22. She wishes to compute a 95% confidence interval.

4. Ecologist will sometimes use tags placed on birds to track and learn about the birds. The tags are typically small metal bands that go around the bird’s leg. Recently ecologist have started to use electronic tags and are interested in learning if there is a difference in the survival rate between the two types of tags. A study looked at the 10-year survival rate of swallows tagged either with a metal tag or an electronic tag. 20% of the 167 metal tagged swallows survived, compared to 36% of the 189 electronic tagged swallows. The margin of error for a 90% confidence interval was found to be 5%. We are interested in a 90% confidence interval for the difference in proportions.

5. 80 first-year students at EKU were randomly assigned roommates. There were 30 students assigned to roommates who brought a video game to college: average GPA after the first semester was 2.84. There were 50 students assigned to roommates who did not bring a video game to college, average GPA after the first semester was 3.105. The margin of error for a 90% confidence interval was found to be .43%. We are interested in a 90% confidence interval for the difference in means.

Answer #1

Solution2:

p^=x/n=20/75=0.2667

Z crit for 90%=1.645

margin of error=Zsqrt(p^(1-p^)/n

=1.645*sqrt(0.2667(1-0.2667)/75

=0.084

this is the reported margin of error

90 % confidence interval for population proprtion

=p^-Z sqrt(p^(1-p^)/n,p^+Z sqrt(p^(1-p^)/n

=0.2667-0.084,0.2667+0.084

=0.1827,0.3507

**lower limit=0.1827**

**upper limit=0.3507**

**Solution3:**

95% confidence interval for true mean

=sample mean-margin of error,sample mean+margin of error

=68.65-3.22,68.65+3.22

=65.43,71.87

**65.43<mu <71.87**

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