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Statistics Worksheet - Page 3

Statistics Worksheet
  • Page 3
 21.  
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.. Approximate the standard error.
a.
4.79
b.
2.5
c.
1.92
d.
0.52


Solution:

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


Correct answer : (4)
 22.  
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. Test at the 0.05 level of confidence and state whether or not to accept the null hypothesis.
a.
The null hypothesis is accepted
b.
The null hypothesis is rejected
c.
Can not be determined
d.
None of the above


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.

97.75 < 98 < 102.25, which is true.
[Sample mean 98.]

The sample mean lies with in the 95% confidence interval.

So, the null hypothesis is accepted.


Correct answer : (1)
 23.  
A standard notebook is supposed to have on an average, 200 pages, with a standard deviation of 2 pages. From a supply of 2500 notebooks, a random sample of 40 notebooks was taken and found that the mean number of pages in the notebook was 197. Determine the standard error.
a.
6.33
b.
2
c.
0.32
d.
3.16


Solution:

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

σe= 240 0.32
[Replace σ = 2 and n = 40.]


Correct answer : (3)
 24.  
A soft drink company claims that the mean quantity of soft drink in a tin is 300 ml with a standard deviation of 5 ml. A random sample of 65 bottles from a shipment had a mean of 296.5 ml. Is the shipment is significantly different from the standard at the 0.05 level of significance?
a.
No
b.
Yes


Solution:

Null hypothesis: Ho: Mean quantity of soft drink in the tin = 300ml
Alternative hypothesis: Ha: Mean quantity of soft drink in the tin ≠ 300ml

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

σe = 565 0.62
[Replace σ = 5 and n = 65.]

μ - 1.96σe = 300 - 1.96 (0.62) 298.78

μ + 1.96σe = 300 + 1.96 (0.62) 301.22

The 95% confidence interval is between 298.78 and 301.22

The sample mean 296.5 does not lie in the confidence interval between 298.78 and 301.22

So, the sample mean does not fall within the 95% confidence interval. So, the null hypothesis is not accepted.

The shipment is significantly different from the standard at the 0.05 level of significance.


Correct answer : (2)
 25.  
A tooth paste tube is supposed to contain 150g of toothpaste, with a standard deviation of 2g. A random sample of 53 tubes is found to have a mean weight of 148.5g. Does this sample differ significantly from the standard at 0.05 level of significance?
a.
Yes
b.
No


Solution:

Null hypothesis:
Ho: Mean weight of toothpaste in the tube = 150g
Alternative hypothesis: Ha: Mean weight of toothpaste in the tube ≠ 150g

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

σe = 253 0.27
[Replace σ = 2 and n = 53.]

μ - 1.96σe = 150 - 1.96 (0.27) 149.47

μ + 1.96σe = 150 + 1.96 (0.27) 150.53

The 95% confidence interval is the interval between 149.47 and 150.53.

The mean weight of the random sample, 148.5 < 149.47

The sample mean does not fall within the 95% confidence interval.

The null hypothesis is not accepted.

The sample differs significantly from the standard at 0.05 level of significance.


Correct answer : (1)
 26.  
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 standard error.
a.
5.0
b.
3.2
c.
1.0
d.
0.18


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.]


Correct answer : (4)
 27.  
The 95% confidence interval for the mean number of tablets in a bottle of calcium tablets is between 149.04 and 150.96. The standard deviation is 3 tablets. Find the size of the random sample that was used to determine this interval.
a.
40
b.
38
c.
6
d.
150


Solution:

The 95% confidence interval is between 149.04 and 150.96.

μ - 1.96σe= 149.04 ---(1)

μ + 1.96σe = 150.96 ----(2)

3.92σe = 1.92
[Subtract equation (1) from equation (2).]

σe 0.49

0.49 = σn
[Standared error, σe= σn.]

0.49 = 3n

n= 6.12

n 37.5 38


Correct answer : (2)
 28.  
The 95% confidence interval for the mean number of mangoes in a carton is between 48.12 and 51.88. If a random sample of 27 cartons were used to determine this interval, what was the standard deviation?
a.
2
b.
3
c.
4
d.
5


Solution:

The 95% confidence interval for the mean number of mangoes in a carton is between 48.12 and 51.88

μ - 1.96σe = 48.12 ---- (1) and μ + 1.96σe = 51.88 ----(2)

3.92 σe = 3.76
[Subtracting (1) from (2).]

σe 0.96

0.96 = σ27
[Standard error, σe= σn.]

σ = 0.96( 27) 5


Correct answer : (4)
 29.  
The 95% confidence interval for the mean weight of a sack of potatoes is between 4.92 and 5.08. If a random sample of 32 sacks was used to determine this interval, what was the standard deviation?
a.
0.04
b.
0.16
c.
3.92
d.
0.23


Solution:

The 95% confidence interval is between 4.92 and 5.08

μ - 1.96σe = 4.92 ----(1) and μ + 1.96σe = 5.08 ----(2)

3.92 σe = 0.16
[Subtracting (1) from (2).]

σe = 0.04

0.04 = σ32
[Standard error, σe= σn.]

σ = 0.04 32 0.23
[Standard deviation.]


Correct answer : (4)
 30.  
The 95% confidence interval for the mean number of sheets of paper in a ream is between 452.76 and 447.24. If the standard deviation is 10 papers, find the size of the random sample that was used to determine this interval.
a.
5.52
b.
50
c.
1.41
d.
7.09


Solution:

The 95% confidence interval is between 447.24 and 452.76.

μ - 1.96σe = 447.24 ----(1) and
μ + 1.96σe = 452.76 ----(2)

3.92σe = 5.52
[Subtract (1) from (2).]

σe 1.41

1.41 = σn
[Standard error, σe = σn.]

n = 10 / 1.41 7.09

n = 50.3 50


Correct answer : (2)

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