Problem 10E

In a sample of 60 electric motors, the average efficiency (in percent) was 85 and the standard deviation was 2.

a. Find a 95% confidence interval for the mean efficiency.

b. Find a 99.5% confidence interval for the mean efficiency.

c. What is the confidence level of the interval (84.63, 85.37)?

d. How many thermostats must be sampled so that a 95% confidence interval specifies the mean to within ±0.35?

e. How many thermostats must be sampled so that a 99.5% confidence interval specifies the mean to within ±0.35?

Solution 10E

Step1 of 6:

Let us consider a random variable X it represents the efficiency of electric motors. Here random variable X follows normal distribution with mean standard deviation and n = 60.

That is

The probability density function of normal distribution is given by

, .

Where,

x = random variable

= mean of X

= variance of X

= standard deviation os X

= mathematical constant and its value is 3.14

Here our goal is:

a). We need to find a 95% confidence interval for the mean efficiency.

b). We need to find a 99.5% confidence interval for the mean efficiency.

c). We need to find the confidence level of the interval (84.63, 85.37).

d). We need to find a 95% confidence interval specifies the mean to within

e). We need to find a 99.5% confidence interval specifies the mean to within

Step2 of 6:

a).

Here we have to find 95% CI, let us take .

Now,

= 0.025

Z-scores(are to the right) is given by

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9750 value falls, it falls in row 1.9 under column 0.06.

Hence,

95% confidence interval for the mean efficiency is given by

(84.4939, 85.5060)

Hence, 95% confidence interval for the mean efficiency is (84.4939, 85.5060).

Step3 of 6:

b).

Here we have to find 99.5% CI, let us take .

Now,

= 0.0025

Z-scores(are to the right) is given by

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9975 value falls, it falls in row 2.8 under column 0.00.

Hence,

99.5% confidence interval for the mean efficiency is given by

(84.2770, 85.7229)

Hence, 99.5% confidence interval for the mean efficiency is (84.2770, 85.7229).

Step4 of 6:

c).

We have and n = 60.

In a given information we have upper bound 85.37

Now,

85.37 - 85 = (0.2581)

0.37 = (0.2581)

=

= 1.4335

Hence, = 1.4335.

Z-scores is given by , and this value is obtained from standard normal table(area under normal curve). In standard normal table we have to see in row 1.4 under column 0.03.

= 0.9236

Area to the right of Z is 1 - 0.9236

= 0.0764.

Now,

= 0.1528

Therefore level is (1-) = 1 - 0.1528

= 0.8472.

Hence level is 0.8472.

Step5 of 6:

d).

We know that , s = 2 .

= 0.35

= 0.35

= 125.44

126

Hence, n = 126.

Step6 of 6:

d).

We know that , s = 2 .

= 0.35

= 0.35

= 256

257

Hence, n = 257.