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Statistics Worksheet
• Page 1
1.
A retail garment shop receives a shipment of trousers, with a standard length of 49.0 in. and a standard deviation of 1.0 inch. A random sample of 32 trousers had a mean length of 48.8 in. Find the 95% confidence interval. a. The interval between 48.65 and 49.35 b. The interval between 32 and 49 c. The interval between 48.8 and 49.9 d. None of the above

Solution:

Standard error, σe= σn, where σ = Standard deviation of the population, n = Number of elements in the sample

σe= 132 0.18
[Replace σ = 1 and n = 32.]

μ - 1.96σe = 49 - 1.96 (0.18) 48.65

μ + 1.96σe = 49 + 1.96 (0.18) 49.35

The 95% confidence interval is the interval between 48.65 and 49.35.

2.
A retail garment shop receives a shipment of trousers, with a standard length of 49.0 in. and a standard deviation of 1.0 inch. A random sample of 32 trousers had a mean length of 48.8 in. Which of the following statements is true? a. The sample mean does not lie with in the 95% confidence interval b. The sample mean lies with in the 95% confidence interval

Solution:

Standard error, σe= σn, where σ = Standard deviation of the population, n = Number of elements in the sample

σe= 132 0.18
[Replace σ = 1 and n = 32.]

μ - 1.96σe = 49 - 1.96 (0.18) 48.65

μ + 1.96σe = 49 + 1.96 (0.18) 49.35

The 95% confidence interval is the interval between 48.65 and 49.35.

48.65 < 48.8 < 49.35, which is true.
[Sample mean is 48.8.]

The sample mean 48.8, lies within the 95% confidence interval.

3.
The mean score on a standardized test is 100 with a standard deviation of 8. A random sample of 48 scores was found to have a mean of 98. Find the standard error. a. 1.15 b. 0.8 c. 8 d. 1

Solution:

σe= σn, where σ is the standard deviation of the population and n is the number of elements in the sample
[Use Standard error formula.]

σe= 848 1.15
[Replace σ = 8, n = 48.]

4.
The mean score on a standardized test is 100 with a standard deviation of 8. A random sample of 48 scores was found to have a mean of 98. Find the 95% confidence interval. a. The interval is between 97.75 and 102.25 b. The interval is between 1.15 and 100 c. The interval is between 100 and 98 d. The interval is between 79.75 and 20.25

Solution:

The 95% confidence interval is between μ - 1.96σe and μ + 1.96σe, where μ is the population mean and σe is the standard error.

σe= σn, where σ is the standard deviation of the population and n is the number of elements in the sample

σe= 848 1.15

μ - 1.96σe = 100 - 1.96 (1.15) 97.75

μ + 1.96σe = 100 + 1.96 (1.15) 102.25

The 95% confidence interval is between 97.75 and 102.25.

5.
To find out how much time is spent at meals by the children of the United States, all the children in a locality of Chicago are interviewed. Which of the statements is true? a. The children in a locality of Chicago represents a random sample of the population for this study b. The children in a locality of Chicago represents the population for this study c. Both of the above d. None of the above

Solution:

The children in a locality of Chicago represents a random sample of the population for this study is true.

6.
In order to find the mean weight of elephants in a zoo, the weights of all the elephants were measured and the mean calculated. State which of the statements is true. a. The group of elephants chosen represents the population for this study. b. The group of elephants chosen represents a random sample of the population for this study. c. Both of the above d. None of the above

Solution:

All the elephants in the zoo are weighed. So they represent the population for this study.

7.
To find the average height of an oak tree, a set of full grown oak trees from a forest were identified and their heights measured. Which of the following statements is true? a. The set of full grown oak trees from a forest represent the population for this study b. The set of full grown oak trees from a forest represent a random sample of the population for this study c. Both are true d. Neither is true

Solution:

The set of full grown oak trees from a forest represent a random sample of the population for this study.

8.
To find out the most popular brand of ice-cream in Florida, all the orders placed for ice cream in an ice-cream parlour in Miami were recorded. Which of the statements is true? a. The orders placed for ice-cream in an ice cream parlour in Miami represents the population for this study b. The orders placed in an ice-cream parlour in Miami does not represent a random sample of the population for this study c. Both of the above d. None of the above

Solution:

The orders placed in an ice cream parlour in Miami represents a random sample of the population for this study.

9.
A company makes wooden wardrobes having a standard height of 55 in., with a standard deviation of 2.5 in. In a random sample of 23 wardrobes, 6 wardrobes measured 58.5 in., 8 wardrobes measured 53.8 in. and 9 wardrobes measured 56 in. State the null hypothesis(Ho) and the alternative hypothesis(H$a$) for the problem. a. Ho: Mean height of a wardrobe = 55.8 in.H$a$ Mean height of a wardrobe ≠ 55.8 in. b. Ho: Mean height of a wardrobe ≠53.8 in.H$a$: Mean height of a wardrobe = 53.8 in. c. Ho: Mean height of a wardrobe ≠55 in.H$a$: Mean height of a wardrobe = 55 in. d. Ho: Mean height of a wardrobe = 55 in.H$a$: Mean height of a wardrobe ≠ 55 in.

Solution:

Mean of random sample = 658.5+853.8+9566 + 8 + 9 = 55.8

Ho: Mean height of a wardrobe = 55 in.
Ha: Mean height of a wardrobe ≠ 55 in.

10.
A company makes wooden wardrobes having a standard height of 55 in., with a standard deviation of 2.5 in. In a random sample of 23 wardrobes, 6 wardrobes measured 58.5 in., 8 wardrobes measured 53.8 in. and 9 wardrobes measured 56 in.. Find the 95% confidence interval. a. The interval is between 53.2 and 56.2 b. The interval is between 53.98 and 56.02 c. The interval is between 0 and 55 d. The interval is betwee 55 and 60

Solution:

The 95% confidence interval is between μ - 1.96σe and μ + 1.96σe.
[Where μ = The population mean.]

Standard error, σe= σn, where σ = Standard deviation of the population, n = Number of elements in the sample.

σe= 2.523
[Replace σ = 2.5 and n = 23.]

σe 0.52

μ - 1.96σe= 55 - 1.96 (0.52) = 53.98

μ + 1.96σe= 55 + 1.96 (0.52) = 56.02

The 95% confidence interval is between 53.98 and 56.02