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Homework #12

Question # 1 Solution

Golf-course designers have become concerned that old courses are becoming obsolete since new technology has given golfers the ability to hit the ball so far. Designers, therefore, have proposed that new golf courses need to be built expecting that the average golfer can hit the ball more than 255 yards on average. Suppose a random sample of 163 golfers be chosen so that their mean driving distance is 255.3 yards, with a standard deviation of 42.8.

Conduct a hypothesis test where H0: \mu = 255 and Ha: mu > 255 by computing the following:

(a) what is the test statistic

(b) what is the p-value

Answer: Test statistic = (255.3 - 255)/(42.8 *√163) = 0.00055 <-— THIS IS YOUR TEST STATISTIC

If you have a TI calculator you can do this function to find your P value

.5-normalcdf(0.00055,0) = .4998005276 <-— THIS IS YOUR P VALUE

Donovan Fonseca

Question # 2

35 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 35 values have a mean of 115 sec and a population standard deviation of 164 sec. Use a .02 significance level to test the claim that the population of all watches has a mean of 0 sec (use a two-sided alternative).

WHAT IS THE TEST STATISTIC AND P-VALUE

115-0/ (164/squareroot(35) ) = 4.14847058 THIS IS THE TEST STATISTIC

.5-normalcdf(0,4.14847058) = .00001674545492 or 1.674545492 X 10 ^-5 <--- THIS IS YOUR P VALUE

Donovan Fonseca

Question # 3 Solution

Find the P-value for the given test statistics:

(a) the test statistics is for a right-tailed test 1.91461035471289

The P-Value is ……..

.5-normalcdf(0,1.91461035471289)

ANSWER: .0277710393

Claculator Function

(b) the test statistics is for a left-tailed test -2.07368911990574

The P-Value is……….

.5-normalcdf(-2.07368911990574,0)

ANSWER: .0190540298

Question # 6

Given the significance level alpha=0.01 find the following:

(a) left-tailed value

Z=?

(b) right-tailed value

Z=?

(c) two-tailed value

Z=?

Solution for A

invNorm(.01)= -2.326347877

Solution for B

invNorm(.99)= 2.32637877

Solution for C

invNorm(.005)= -2.575829303 ---> 2.575829303 TAKE ABSOLUTE VALUE

### Quiz - 1

A study is being conducted of how far UTPA students live from campus. Survey takers will be in the parking lot by the baseball field between 8 and noon on Tuesday asking students to take the poll.

- What is the population?

UTPA students

- What is the sample?
- What is the variable of interest?
- What type of variable is it?
- What are the possible biases of this sample, if any?

### Example 1

Solve for mean, sample variance and standard deviation to the following data:

0, 1,3,6,6,8,9,11,13,17

Mean= (0+1+3+6+6+8+9+11+13+17)/ 10

=7.4

Sample variance:

7.4^2

6.4^2

3.4^2

3.4^2

.6^2

1.6^2

3.6^2

5.6^2

9.6^2

246.68/(n-1)

=246.68/9

=27.41

Standard Deviation:

(square root) of 27.41= 5.24

- jigoodship

### Example 2

Quiz 6

Tim 65 % chance of making free throws (.35 chance in failing)

Sarah 85% chance of making free throws (.15 chance in failing)

1. Probability of each makeing 2 free thows in a row

Tim: .65*.65=0.4225

Sarah: .85*.85=0.7225

2. Tim makes at least 2 out of 3 attempts

SSF= .65*.65*.35= 0.147875

SSS= .65*.65*.65= 0.274625

SFS= .65*.35*.65= 0.147875

FSS= .35*.65*.65= 0.147875

Total= 0.71825

3. Each make one together

.65*.85= .5525

4. Sarah at least 2 out of 3 attempts

SSS= .85*.85*.85= 0.614125

SSF= .85*.85*.15= 0.108375

SFS= .85*.15*.85= 0.108375

FSS= .15*.85*.85= 0.108375

Total= 0.93925

-JIGOODSHIP

## Chapter 2 Example :

1.Find the mean and the standard deviation 4,8,9,8,6,5,7,5,8

Mean =60/9=6.67

variance=(2.67^2+1.67^2+1.67^2+0.67^2+0.33^2+1.33^2+1.33^2+1.33^+2.33^2)/9=2.21

Standard deviation=/2.21=1.49

2.Chebyshev's theorem gurantee that what proportion of a distributio will be included between the follwing:

(a)X -2s and X+2s

=75%

(b) x-3s and X+3s

=89%

3.

The average clean up time for a crew for a medium size firm is 84 hrs(X) and the standard deviation(S) is 6.8 hrs. Assume empirical rule is appropiate. Within what interval will the total cleanup time fall 95% of the time.

84-2(6.8) to 84+2(6.8) =70.4hrs to 97.6 hrs

-m

Probability question:

1) Explain why this data is incorrect. P(A)= 0.2 P(B)= 0.7 P(A and B)= 1.0

Answer: because P(A and B)= is bigger than P(A) and P(B). There is an entire overlap of the other.

- jigoodship

Example using tree:

Three kids were each given a reversible jacket; one side was solid black and the other side was checkered. What are the chances that two of them wore the checkered side,C, and one of them the black side,B?

Solution: Answer:

C * 3/8 have 2 checkered and one black

C< B or 37.5%

B<

B< C

B

C

C < B *

C<

B< C *

B

-Samantha Ambriz