Question

A random sample of 44 taxpayers claimed an average of ​$9,700 in medical expenses for the...

A random sample of 44 taxpayers claimed an average of ​$9,700 in medical expenses for the year. Assume the population standard deviation for these deductions was ​$2,379. Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below.

a.5%

b.10%

c.20%

a. The confidence interval with a 5% level of significance has a lower limit of ​$ nothing and an upper limit of ​$ nothing .

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The caloric consumption of 44 adults was measured and found to average 2,117. Assume the population...
The caloric consumption of 44 adults was measured and found to average 2,117. Assume the population standard deviation is 251 calories per day. Construct confidence intervals to estimate the mean number of calories consumed per day for the population with the confidence levels shown below. a. 93 % b. 96 % c. 98 % a. The 93 % confidence interval has a lower limit of and an upper limit of
The average selling price of a smartphone purchased by a random sample of 44 customers was...
The average selling price of a smartphone purchased by a random sample of 44 customers was ​$299. Assume the population standard deviation was ​$33. a. Construct a 95​% confidence interval to estimate the average selling price in the population with this sample. b. What is the margin of error for this​ interval? a. The 95​% confidence interval has a lower limit of ​$?? and an upper limit of $??. ​(Round to the nearest cent as​ needed.) b. The margin of...
The average selling price of a smartphone purchased by a random sample of 41 customers was...
The average selling price of a smartphone purchased by a random sample of 41 customers was ​$323. Assume the population standard deviation was ​$33. a. Construct a 90​% confidence interval to estimate the average selling price in the population with this sample. b. What is the margin of error for this​ interval? a. The 90% confidence interval has a lower limit of ​$ nothing and an upper limit of ​$ nothing .
The caloric consumption of 37 adults was measured and found to average 2 comma 199. Assume...
The caloric consumption of 37 adults was measured and found to average 2 comma 199. Assume the population standard deviation is 259 calories per day. Construct confidence intervals to estimate the mean number of calories consumed per day for the population with the confidence levels shown below. a. 91 % b. 96 % c. 97 % a. The 91 % confidence interval has a lower limit of nothing and an upper limit of ____. ​(Round to one decimal place as​...
The caloric consumption of 38 adults was measured and found to average 2,136. Assume the population...
The caloric consumption of 38 adults was measured and found to average 2,136. Assume the population standard deviation is 270 calories per day. Construct confidence intervals to estimate the mean number of calories consumed per day for the population with the confidence levels shown below. a.92% b.95% c.99% a. The 92% confidence interval has a lower limit of and an upper limit of .
A random sample of 20 statistics examinations was taken. The average score, in the sample, was...
A random sample of 20 statistics examinations was taken. The average score, in the sample, was 84 with a standard deviation of 3.5. Develop a 95% confidence interval for the average examination score of the population of the examinations. a. The critical value for the confidence interval   (Provide 3 decimals) b. Lower limit   (Provide 2 decimals) c. Upper limit   (Provide 2 decimals)
A random sample of 328 medical doctors showed that 162 had a solo practice. (a) Let...
A random sample of 328 medical doctors showed that 162 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.) (b) Find a 95% confidence interval for p. (Use 3 decimal places.) lower limit upper limit Give a brief explanation of the meaning of the interval. (Choose one of the below) 5% of the all confidence intervals would include the true...
A random sample of 15 college​ men's basketball games during the last season had an average...
A random sample of 15 college​ men's basketball games during the last season had an average attendance of 5,154 with a sample standard deviation of 1,775. Complete parts a and b below. a. Construct a 99​% confidence interval to estimate the average attendance of a college​ men's basketball game during the last season. The 99​% confidence interval to estimate the average attendance of a college​ men's basketball game during the last season is from a lower limit of to an...
A random sample of 324 medical doctors showed that 166 had a solo practice. (a) Let...
A random sample of 324 medical doctors showed that 166 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.) (b) Find a 99% confidence interval for p. (Use 3 decimal places.) lower limit upper limit Give a brief explanation of the meaning of the interval. 1% of the all confidence intervals would include the true proportion of physicians with solo...
A random sample of 336 medical doctors showed that 160 had a solo practice. (a) Let...
A random sample of 336 medical doctors showed that 160 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.) ------------ (b) Find a 99% confidence interval for p. (Use 3 decimal places.) lower limit       upper limit Give a brief explanation of the meaning of the interval. 1% of the confidence intervals created using this method would include the...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT