98. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce. a. i. x ¯ = ii. σ = iii. sx = b. In words, define the random variable X. c. In words, define the random variable X ¯ . d. Which distribution should you use for this problem? Explain your choice. e. Construct a 90% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. Construct a 98% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. g. In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e. h. In complete sentences, give an interpretation of what the interval in part f means.

The sample mean is denoted by x bar and standard deviation by sx and it is the mean and standard deviation of the sample not the population from which the sample is chosen.

The data given in problem answers the first part.

The mean weight of the sample denoted by x bar x ¯ was two ounces with a standard deviation denoted by sx of 0.12 ounces.

The population standard deviation is denoted by σ known to be 0.1 ounce.

a. i. x ¯ =___2_____ ii. σ =___0.1_____ iii. sx =___0.12_____

e. Sub part i) 1.959 ; 2.041

Sub part iii) The error bound is 0.041

f. Sub Part i) 1.935 ; 2.065

Sub Part iii) The error bound is 0.0645

Part b)

The random variable X is any number defining the weights of the sample chosen from the population.

Part c)

The random variable X bar is the average weight of the sample obtained by adding the total weights of the sample divided by the number of the sample size.

Part d)

The normal distribution is used to solve the problem because the sample size is less than 30.

Part e

The 90 % confidence interval is calculated as ∝ = 1- 90%

∝ = 0.1

∝/2= 0.1/2= 0.05

xbar± z(∝/2) σ /√n

Putting the values

2± 1.645 (0.1)/√16

2± 0.041

or

Sub part i) 1.959 ; 2.041

Sub part iii) The error bound is 0.041

Part f

The 98 % confidence interval is calculated as ∝ = 1- 98%

∝ = 0.02

∝/2= 0.1/2= 0.01

xbar± z(∝/2) σ /√n

Putting the values

2± 2.58 (0.1)/√16

2± 0.0645

or

Sub Part i) 1.935 ; 2.065

Sub Part iii) The error bound is 0.0645

Part g

The confidence interval in part f is larger than the confidence interval in part e because we are 98 percent confident about the accuracy of the result. The interval is 0.01 for two tailed test which is very less as compared to 0.05 for 90% confidence interval.

The larger the area of the confidence interval less sure we are about the accuracy of the result hence getting a smaller value .

Part h

On the average 90 % of the 16 sample will have the true value of the mean.

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