Question

1. In a survey, 20 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $43.1 and standard deviation of $9.4. Construct a confidence interval at a 90% confidence level.

Express your answer in the format of ¯xx¯ ±± Error.

$ ±± $

2. The body temperatures in degrees Fahrenheit of a sample of 5 adults in one small town are:

97.7

98.4

96.4

96.7

99.2

Assume body temperatures of adults are normally distributed. Based on this data, find the 80% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

80% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

3.

The body temperatures in degrees Fahrenheit of a sample of adults in one small town are:

98.8

97.5

99.5

99.8

98

96.9

97.4

99.6

99

99.4

Assume body temperatures of adults are normally distributed. Based on this data, find the 98% confidence interval of the mean body temperature of adults in the town. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places. Assume the data is from a normally distributed population.

98% C.I. =    

4.

You are interested in finding a 90% confidence interval for the average commute that non-residential students have to their college. The data below show the number of commute miles for 14 randomly selected non-residential college students.

25

9

9

15

9

25

17

16

14

27

13

23

7

7

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean commute for non-residential college students is between  and   miles.

c. If many groups of 14 randomly selected non-residential college students are surveyed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of commute miles and about  percent will not contain the true population mean number of commute miles.

5.

A psychiatrist is interested in finding a 90% confidence interval for the tics per hour exhibited by children with Tourette syndrome. The data below show the tics in an observed hour for 15 randomly selected children with Tourette syndrome. Round answers to 3 decimal places where possible.

1

8

7

4

10

9

2

9

10

3

11

11

12

6

6

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean number of tics per hour that children with Tourette syndrome exhibit is between  and .

c. If many groups of 15 randomly selected children with Tourette syndrome are observed, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of tics per hour and about  percent will not contain the true population mean number of tics per hour.

6.

You are interested in finding a 90% confidence interval for the mean number of visits for physical therapy patients. The data below show the number of visits for 13 randomly selected physical therapy patients.

17

26

20

26

18

15

8

22

5

19

27

24

28

a. To compute the confidence interval use a ? t z  distribution.

b. With 90% confidence the population mean number of visits per physical therapy patient is between  and   visits.

c. If many groups of 13 randomly selected physical therapy patients are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population mean number of visits per patient and about  percent will not contain the true population mean number of visits per patient.

Answer 1

(1)

df = 20 - 1 = 19

= 0.10

From Table, critical values of t = 1.729

So,

Confidence Interval:

So,

Answer is:

$43.1 $3.634

(2)

From the given data,the following statistics are calculated:

n = 5

= 97.68

s = 1.1649

df = 5 - 1 = 4

= 0.20

From Table, critical values of t = 1.533

So,

Confidence Interval:

So,

Answer is:

(96.881,98.479)

(3)

From the given data,the following statistics are calculated:

n = 10

= 98.59

s = 1.0535

df = 10 - 1 = 9

= 0.02

From Table, critical values of t = 2.821

So,

Confidence Interval:

So,

Answer is:

(97.65,99.53)

(4)

(a)

To compute the confidence interval use a t   distribution.

(b)

From the given data,the following statistics are calculated:

n = 14

= 15.4286

s = 7.0788

df = 14 - 1 = 13

= 0.10

From Table, critical values of t = 1.771

So,

Confidence Interval:

So,

Answer is:

(12.078,18.779)

(c)

. About 90 % percent of these confidence intervals will contain the true population mean number of commute miles and about 10 % percent will not contain the true population mean number of commute miles.

(5)

(a)

To compute the confidence interval use a t distribution.

(b)

From the given data,the following statistics are calculated:

n = 15

= 7.2667

s = 3.4942

df = 15 - 1 = 14

= 0.10

From Table, critical values of t = 1.761

So,

Confidence Interval:

So,

Answer is:

(5.678,8.856)

(c)

. About 90 % percent of these confidence intervals will contain the true population mean number of commute miles and about 10 % percent will not contain the true population mean number of commute miles.

(6)

(a)

To compute the confidence interval use a t distribution.

(b)

From the given data,the following statistics are calculated:

n = 13

= 19.6154

s = 7.1360

df = 13 - 1 = 12

= 0.10

From Table, critical values of t = 1.782

So,

Confidence Interval:

So,

Answer is:

(16.088,23.143)

(c)

. About 90 % percent of these confidence intervals will contain the true population mean number of commute miles and about 10 % percent will not contain the true population mean number of commute miles.