exam at the imapr The sd we in the exam sval for the mathematics skills, a simpl
ID: 3054949 • Letter: E
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exam at the imapr The sd we in the exam sval for the mathematics skills, a simple random sample of The students were given an 4. In order to test the effectiveness of a program to 45 fifth graders was chosen to participate in the program. beginning and again at the end of the program. points, with a sample standard deviation of 4.7 points. Construct a increase in score. The developers of the program claim it points. Does the confidence interval contradict this claim? Explain. Type of problem: Other information requested: 99% confidence interval for the mean ea increase in the exam score was 12.2 opers of the program claim it will produce a mean increase o A 1989 sample of 130 college women who visited a gynecologist at a particular university in a northeasterm U.S. indicated that 113 were sexually experienced. Assuming that these women were a simple random sample from the population of all women at that university, calculate a 95% confidence interval for the proportion of the population who are sexually active Type of problem: Other information requested: 5. a. Do you think it is reasonable to assume that these women form a random sample? Explain your answer b. Would the interval have been wider, narrower, or the same width if 520 women had been samples? You don't need to perform the calculation, just briefly explain your answer. 6. A questionnaire of spending habits was given to a random sample of college students. Each student was asked to record and report the amount of money they spent on textbooks in a semester. The 130 students resulted in an average of $422 with a standard deviation of $57. a Give a 90% confidence interval for the mean amount of money spent by college students on sample of (2 points) textbooks. Is it true that 90% of the students spent the amount of money found in the interval fron part (a)? Explain your answer bExplanation / Answer
M = 12.2
t = 2.69
sM = ?(4.72/45) = 0.7
? = M ± t(sM)
? = 12.2 ± 2.69*0.7
? = 12.2 ± 1.886
You can be 99% confident that the population mean (?) falls between 10.314 and 14.086.
Since 15 does not fall in the confidence interval, the confidence interval contradicts the claim
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