# Statistics Worksheet - Page 2

Statistics Worksheet
• Page 2
11.
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.. Test at 0.05 level of confidence and find out if the null hypothesis can be accepted.
 a. Cannot be determined b. The null hypothesis is not accepted c. The null hypothesis is accepted

#### 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

The mean height of the samples = (6 × 58.5) + (8 × 53.8) + (9 × 56)23

= 351 + 430.4 + 50423 55.89

53.98 < 55.89 < 56.02, which is true.
[Sample mean is 55.89.]

Since the sample mean falls within the 95% confidence interval, the null hypothesis is accepted.

12.
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. State the null hypothesis(Ho) and the alternative hypothesis(Ho).
 a. Ho: The mean number of pages in a notebook ≠ 200, H$a$: The mean number of pages in a notebook = 200 b. Ho: The mean number of pages in a notebook = 200, H$a$: The mean number of pages in a notebook ≠ 200 c. Ho: The mean number of pages in a notebook ≠197, H$a$: The mean number of pages in a notebook = 197 d. Ho: The mean number of pages in a notebook = 197, H$a$: The mean number of pages in a notebook ≠ 197

#### Solution:

Ho: The mean number of pages in a notebook = 200
Ha: The mean number of pages in a notebook ≠ 200

13.
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 95% confidence interval.
 a. The interval is between 197 and 200 b. The interval is between 0 and 197 c. The interval is between 0 and 200 d. The interval is between 199.37 and 200.63

#### 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.]

μ - 1.96σe = 200 - 1.96 (0.32) » 199.37

μ + 1.92σe = 200 + 1.96 (0.32) » 200.63

The 95% confidence interval is between 199.37 and 200.63

14.
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. Test at the 0.05 level of confidence and find out if the null hypothesis can be accepted.
 a. The null hypothesis is not accepted b. Cannot be determined c. The null hypothesis is accepted

#### 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.]

μ - 1.96σe = 200 - 1.96 (0.32) » 199.37

μ + 1.92σe = 200 + 1.96 (0.32) » 200.63

The 95% confidence interval is between 199.37 and 200.63

197 does not lie within the confidence interval between 199.37 and 200.63
[Sample mean is 197.]

The null hypothesis is not accepted.

15.
Radio tubes manufactured by a company are expected to have a mean lifetime of 400 hours, with a standard deviation of 30 hours. Random samples of 10 tubes were tested. It was found that 3 tubes had a lifetime of 398 hours, 5 of them had a lifetime of 410 hours and 2 had a lifetime of 400 hours. Is this case significantly different from the standard at the 0.05 level of significance?
 a. Can not be determined b. Yes c. No

#### Solution:

Null hypothesis: Ho: Mean life time of the radio tubes = 400 hours
Alternative hypothesis: Ha: Mean life time of the radio tubes ≠ 400 hours

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

σe = 3010 9.49
[Replace σ = 30 and n = 10.]

μ - 1.96σe = 400 - 1.96 (9.49) 381.4

μ + 1.96σe = 400 - 1.96 (9.49) 418.6

The 95% confidence interval lies between 381.4 and 418.6

Sample mean = (3 × 398) + (5 × 410) + (2 × 400)10 = 404.4

381.4 < 404.4 < 418.6, which is true.
[Sample mean is 404.4]

The sample mean falls within the 95% confidence interval. So, the null hypothesis is accepted.
Hence, this case is not significantly different from the standard at the 0.05 level of significance.

16.
A retail garment shop receives a shipment of trousers, with a standard length of 49.0 in. and a standard deviation of 1.0 in. A random sample of 32 trousers had a mean length of 48.8 in. State the null hypothesis(Ho) and alternative hypothesis(H$a$).
 a. Ho: Mean length of trousers = 32.0, H$a$: Mean length of trousers ≠ 32.0 b. Ho: Mean length of trousers = 49.0, H$a$: Mean length of trousers ≠ 49.0 c. Ho: Mean length of trousers = 48.8, H$a$: Mean length of trousers ≠ 48.8 d. None of the above

#### Solution:

Ho: Mean length of trousers = 49.0
Ha: Mean length of trousers ≠ 49.0

17.
State the null hypothesis and alternative hypothesis for the situation: The mean number of hours spent on practice and exercise per day by baseball players is 6, with a standard deviation of $\frac{1}{2}$ hour. A random sample of 10 players in a baseball team was taken and found that the number of hours spent on practice and exercise per day was 5 hours.
 a. Ho: The mean number of hours spent on practice and exercise = 5 hours, H$a$: The mean number of hours spent on practice and exercise ≠ 5 hours b. Ho: The mean number of hours spent on practice and exercise = 6 hours, H$a$: The mean number of hours spent on practice and exercise ≠ 6 hours c. Ho: The mean number of hours spent on practice and exercise ≠ 5 hours, H$a$: The mean number of hours spent on practice and exercise = 5 hours d. Ho: The mean number of hours spent on practice and exercise ≠ 6 hours, H$a$: The mean number of hours spent on practice and exercise = 6 hours

#### Solution:

Ho: The mean number of hours spent on practice and exercise = 6 hours.
Ha: The mean number of hours spent on practice and exercise ≠ 6 hours.

18.
The mean life span of an electric bulb is 350 hours, with a standard deviation of 10 hours. A random sample 40 of a particular brand of electric bulbs is taken and found that their mean life span was 348 hours. State the null hypothesis( Ho) for the situation.
 a. Ho: The mean life span of the bulbs = 400 b. Ho: The mean life span of the bulbs = 300 c. Ho: The mean life span of the bulbs = 450 d. Ho: The mean life span of the bulbs = 350

#### Solution:

Ho: The mean life span of the bulbs = 350 hours.

19.
The mean number of candies in a packet is 175, with a standard deviation of 5 candies. A random sample of 70 packets from a certain shipment has a mean of 174.2 candies. Which of the following statements is true?
1.The null hypothesis is accepted.
2.Standard error = 1.78
3.The shipment is significantly different from the standard at 0.05 level.
4.The 95% confidence interval lies between 173.8 and 176.2
 a. 2 and 3 are true b. 1 and 4 are true c. All are true d. All are false

#### Solution:

Null hypothesis: Ho: The mean number of candies in a packet = 175
Alternative hypothesis: Ha: The mean number of candies in a packet ≠175

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

σe = 570 0.6 Statement 2 is false.
[Replace σ = 5 and n = 70.]

μ - 1.96σe = 175 - 1.96 (0.6) 173.8

μ + 1.96σe= 175 + 1.96 (0.6) 176.2

The 95% confidence interval lies between 173.8 and 176.2 statement 4 is true.

173.8 < 174.2 < 176.2, which is true. So, 174.2 lies in the 95% confidence interval.

Therefore, the null hypothesis is accepted statement 1 is true.

So, the shipment is not significantly different from the standard at 0.05 level. The statement 3 is false.

The correct choice is B.

20.
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. State the null hypothesis and alternative hypothesis.
 a. Ho: The mean score = 100, H$a$: The mean score ≠ 100 b. Ho: The mean score = 8, H$a$: The mean score ≠ 52 c. Ho: The mean score = 98, H$a$: The mean score ≠ 98 d. Ho: The mean score = 48, H$a$: The mean score ≠ 48

#### Solution:

Ho: The mean score = 100
Ha: The mean score ≠ 100.