6.5.2. Confidence interval for : small samples

Let us turn our attention to finding confidence interval for when sample size is small. Using

We can derive the formula for confidence interval for the case when a small sample is selected from a normally distributed population with mean and unknown variance. It is given by

where is the number for which

The random variable has a Student’s t distribution with degrees of freedom. (Fig. 6.10).

Remark:

If the sample is available, then standard deviation can be calculated as

, where or

Example:

For the t distribution with n =10, find the number b such that

S olution:

The probability in the interval

is 0.80. (Fig. 6.11).

We must have a probability of 0.10 to the right of b and a probability of 0.10 to the left

of –b.

So

and .

Example:

A random sample of 25 busses shows a sample mean of 225 passengers carried per day per bus. The sample standard deviation is computed to be 60 passengers. Find a 90% confidence interval for the mean number of passengers carried per bus during a 1 –day period.

Solution:

A 90 % confidence interval for the mean is given by

, so

=90 %

and

After substitution we obtain

or .

We are 90 % confident the mean number of passengers carried per day by bus is between 204.5 and 245.5, because 90 % of the intervals calculated in this manner will contain the true mean number of passengers carried per day per bus.

Exercises

1. In each case, find the number b so that

a) when n =7

b) when n =16

c) when n =9

d) when n =12

2. For each of the following, find the area in the appropriate tail of the t distribution

a)

b)

c)

d)

3. Find the value of t from t – distribution table for each of the following:

a) Confidence level = 99 % and d. f . =18

b) Confidence level = 95 % and n =26

c) Confidence level = 90 % and d. f . =15

4. The mean number of the sample of 25 bolts produced on a specific machine per day was found to be 47 with a standard deviation of 2.4. Assume that the number of bolts produced per day on this machine has a normal distribution.

Construct a 90 % confidence interval for the population mean .

5. A random sample of 16 cars, which were tested for fuel consumption, gave a mean of 26.4 miles per gallon with a standard deviation of 2.3 miles per gallon. Assuming that the miles per gallon given by cars have a normal distribution, find a 99 % confidence interval for the population mean .

6. A sample of eight adults was taken, and these adults were asked about the time they spend per week on sport activities. Their responses (in hours) are as follows:

45; 12; 31; 16; 28; 14; 18; 26

Make a 95 % confidence interval for the mean of time spent per week by all adults on sport activities.

7. A sample of 10 customers who visited a supermarket was taken. The following data give the money (in dollars) they spent during that visit:

74; 89; 121; 63; 146; 47; 91; 28; 84; 76

Assuming that the money spent by all customers at this supermarket has a

normal distribution, construct a 90 % confidence interval for the population mean.

8. The mean time taken to design a home plan by 20 designers was found to be 185 minutes with a standard deviation of 23 minutes. Assume that the time taken by all designers this home plan is normally distributed.

Construct a 99 % confidence interval for the population mean .

Answers

1. a) 1,943; b) 2.131; c) 2.896; d) -2.718; 2. a) ; right tail;

b) ; left tail; c) less than ; left tail; d) less than ; right tail; 3. a) 2.878; b) 2.060; c) 1.753; 4. a) (46.18; 47.82);

5. (24.71; 28.09); 6. (14.52; 32.98); 7. 62.24 to 101.56; 8. (170.29; 199.71).