Questions 1 through 7 Suppose SAT math scores follow a Normal distribution with

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Questions 1 through 7 Suppose SAT math scores follow a Normal distribution with -100. An SRS of 100 students gave an average score of 512, Find a 99% confidence interval for the average SAT score. 2. A survey of 30 children found that 11 preferred cake to ice cream. Find a 95 % confidence interval for the proportion of all children who prefer cake to ice cream. 3. An SRS of 60 households in Greenville is taken. In the sample, there are 45 households that decorate their houses with lights for the holidays. A simple random sample of 50 households is also taken from the neighboring town of Brownsboro. In the same, there are 40 households that decorate their houses. (a) What is the standard error of the diference in sample p roportions? (Use Greenville as the first group and Brownsboro as the second) (b) Find a 95% confidence interval for the difference in population proportions of households that decorate their houses with lights for the holidays. 4 You want to find a 97% confidence interval for the rnean heights of USA students with a margin of error of 1 inch. You know the standard deviation of heights is 2.4 inches. What sample size should you use? 5 Jill collects a large sample and finds a 95% confidence interval for the mean numbers of hours USA students spend on Facebook per day to be [.75,2.05]. Construct a 99% confidence interval from, the same data. 6. An SRS of 128 Republicans gave a mean age of 39 years and standard deviation of 8 years. An SRS of 200 Democrats gave a mean age of 40 years and a standard deviation of 10 years. Find a 90% confidence interval for the difference in average age of republicans and democrats. 7. Many college students change their major several times throughout their college career. A survey found that 1230 out of a total of 3401 students reported changing their major at least twice. Find an estimate of the true proportion of students who change their major at least twice with a 99% confidence interval.

Explanation / Answer

1

we know that the confidence interval is given as

mean +- z*sd/sqrt(n)

so mean = 512
sd = 100
n = 100
z = 2.58 , from the z table for 99% CI
putting the values in the equation

512 +- 2.58*100/sqrt(100), thus solving for plus and sign , we get the confidence interval as

upper limit = 537.8
lower limit = 486.2

2

CI of proportion is given as

p+-z*sqrt(p*(1-p)/n)

for 95% , z = 1.96 from the z tables
p = 11/30 = 0.36

so putting the values

0.36 +-1.96*sqrt(0.36*(1-0.36)/30)

lower limit = 0.19
upper limit = 0.53

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