The sugar content in a one-cup serving of a certain breakfast cereal was measured for a sample of 140 servings. The average was 11.9 g and the standard deviation was 1.1 g.

a. Find a 95% confidence interval for the mean sugar content.

b. Find a 99% confidence interval for the mean sugar content.

c. What is the confidence level of the interval (11.81, 11.99)?

d. Howlarge a sample is needed so that a 95% confidence interval specifies the mean to within ±0.1?

e. Howlarge a sample is needed so that a 99% confidence interval specifies the mean to within ±0.1?

Solution 11E

Step1 of 6:

Let us consider a random variable X it represents the sugar content contained in a certain breakfast. Here random variable X follows normal distribution with mean standard deviation and n = 140.

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 sugar content.

b. We need to find a 99% confidence interval for the mean sugar content.

c. We need to find confidence level of the interval (11.81, 11.99).

d. We need to find a 95% confidence interval specifies the mean to within ±0.1.

e. We need to find a 99% confidence interval specifies the mean to within ±0.1.

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 sugar content is given by

(11.7177, 12.0822)

Hence, 95% confidence interval for the mean sugar content is (11.7177, 12.0822).

Step3 of 6:

b).

Here we have to find 99% CI, let us take .

Now,

= 0.005

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 0.9950 value falls, it falls in row 2.5 under column 0.07.

Hence,

99% confidence interval for the mean sugar content is given by

(11.6620, 12.1380)

Hence, 99% confidence interval for the mean sugar content is (11.6620, 12.1380).

Step4 of 6:

c).

We have and n = 140.

In a given information we have upper bound 11.99

Now,

11.99 - 11.9 = (0.0929)

0.09 = (0.0929)

=

= 0.9687

Hence, = 0.9687.

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 0.9 under column 0.06.

= 0.8315

Area to the right of Z is 1 - 0.8315

= 0.1685.

Now,

= 0.3370

Therefore level is (1-) = 1 - 0.3370

= 0.6630.

Hence level is 0.6630.

Step5 of 6:

d).

We know that , s = 1.1 .

= 0.1

= 0.1

= 3.8416

4

Hence, n = 4.

Step6 of 6:

e).

We know that , s = 1.1 .

= 0.1

= 0.1

= 6.6049

7

Hence, n = 7.