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Christian Rivero

Statistics Solution

# Problem Set 2

1. Light Bulbs problemFour brands of light bulbs are being considered for use in the final assembly area of the Saturn plant in

Spring Hill, Tennessee. The director of purchasing asked for samples of 100 from each manufacturer. The

numbers of acceptable and unacceptable bulbs from each manufacturer are shown below. At the .05

significance level, is there a difference in the quality of the bulbs?

Manufacturer

A B C D

Unacceptable 12 8 5 11

Acceptable 88 92 95 89

Total 100 100 100 100

Answer:

Null: There is no difference in the quality of bulbsAlternative: There is difference in the quality of bulbs

Uses chi square test to test the hypothesis that there a difference in the quality of the bulbs produced bydifferent manufacturers.

between A and B: nobetween A and C: yesbetween A and D: no

between B and C: nobetween B and D: nobetween C and D: yes

p=157 >a =0.05 accept the null

Answer: Accept Null Hypothesis. There is no difference in the quality of the bulb produced.


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2. Quality control department town problem ( Same problem as Number 10)National News Sports

Undercharge 20 10Overcharge 15 30

Correct Price 200 225Total 235 265


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3. DPWH ProblemAnswer:

Null Hypothesis: There is no significant relationship between the number of bidders and the amount ofwinning bid on highway projects.

Alternative Hypothesis: There is a significant relationship between the number of bidders and the amount

of winning bid on highway projects.

An inverse relation has a regression equation ofy = 5.72 + 13.87/x 2.3707

(r = 0.58757)A straight-line regression has the formulay = 11.23 - 0.4667x 2.0738

(r = -0.70638), which is a slightly better fit, but neither is a particularly good fit, as shown by the rmserrors and the correlation factors

Each indicates a decrease in bid with an increase in bidders.

Decision: Reject the Null (p=0.70). There is a significant positive relationship between the number ofbidders and the amount of winning bid on highway projects.


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4. Cardio Glide

Answer:Person Months owned(X) Hours exercised(Y) X* X Y*Y X*YJim 12 4 144 16 48

Claire 2 10 4 100 20Juan 6 8 36 64 48Neil 9 5 81 25 45Jonalyn 7 5 49 25 35William 2 8 4 64 16Joshua 8 3 64 9 24Melvin 4 8 16 64 32

John 10 2 100 4 20James 5 5 25 25 25

Total 65 58 523 396 313

The hypothesis to be tested is:

Ho: r = 0H1: r < 0 (hypothesizing a significant negative correlation between the two variables - a one tailed test)

The test statistic is,. t = r [(n-2)/(1-r2 )]= - 0.8269 * [(10-2)/(1-(- 0.8269)2 )]

= - 4.1593From the t-table for 0.01 level of significance and for 10-2 = 8 degrees of freedom we get the criticalvalue = 2.896

Decision: Since the numerical value of the test statistic is greater than the critical value we reject the nullhypothesis with 99% confidence. Hence we conclude that there is a negative association between thenumber of months since the glide was purchased and the length of time the equipment was used lastweek.


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5. home sample

A sample of 12 homes sold last week is selected. Can we conclude that as the size of the home (reportedbelow in thousands of square feet) increases, the selling price (reported in thousands) also increases?

Home Size Selling

(thousands Price

of square feet) ( thousands)

X Y

1.4 100

1.3 110

1.2 105

1.1 120

1.4 80

1.0 105

1.3 110

0.8 85

1.2 105

0.9 75

1.1 70

1.1 95

Total 13.8 1160

Null Hypothesis: There is no significantrelationship between the size of the home andthe selling price

Alternative Hypothesis: There is a significantrelationship between the home size and theselling price

Level of Significance: 0.05

H1 : p0; p>0. Reject H 1 if t >

t =0076.1

212087.

= .275

Decision: Accept the null (p=.30) There is nosignificant relationship between the size of thehome and the selling price.

We cannot conclude that there is a positivecorrelation between the size of thehome and the selling price.

Thus, there is no association between the homesize and the selling price of 12 homes sold


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6. Academic Psychologist Problem

* Using SPSS

ANOVA

Frequency

Sum of Squares df Mean Square F Sig.

Between Groups 3168.167 2 1584.083 7.366 .002

Within Groups 7096.583 33 215.048

Total 10264.750 35

Post Hoc Tests

Homogeneous Subsets

Frequency

Tukey Ba

esteem N

Subset for alpha = 0.05

1 2

low 12 14.83

medium 12 21.67

high 12 37.25

Means for groups in homogeneous subsets are

displayed.

a. Uses Harmonic Mean Sample Size =

12.000.


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7. Body Esteem Problem

Answer

Null Hypothesis: There is no significant difference among participants with high, medium and low

body esteem in terms of high frequency of sexual intercourse.

Alternative: There is a significant difference among participants with high, medium and low body

esteem in terms of high frequency sexual intercourse

Level of significance: 0.05

Computation: ANOVA single factor

Summary

Groups Count Sum Average VarianceHigh BodyEsteem

12 447 37.25 3777.4773

Medium BodyEsteem

12 260 21.66667 157.8788

Low Body 12 178 14.83333 109.789

Anova

Source ofVariation

SS df MS F P-value F crit

Betweengroups

3168.167 2 1584.083 7.366186 0.002265 3.2849

Withingroups

7096.583 33 215.048

Total 10264.75 35

Decision: (F crit 3.2,F7.3) Reject the null. Therefore, there is a significant difference among

participants whos body esteem is high, medium and low in terms of sexual intercourse.


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8. Survey conducted at a University for Filipino Food

Answer:

Null: There is no significant relationship with the richer Filipino food with the increase in weight

Alternative: There is a significant relationship with the richer Filipino food with the increase in weight

It is most appropriate to adopt a paired t-test for this problem. Hence the 2 sets of data will be

combined and tested based on their differences. There is no need to be concerned with equal or

unequal variances.

If X1 and X2 denote the weight of a student before and after, respectively, D = X2 - X1 shall denote

the difference. It is assumed that X1 and X2 are normally distributed.

Based on the data, we obtain the following results

Sample size n = 11

Sample mean D-bar = 7.364

Sample standard deviation sD = 8.370

The hypotheses are

Ho: 1 = 2H1: 1 2

The test statistic is

T = D-bar / sD/n = 7.364 / 8.370/11 = 2.918

Given that = 0.01, the critical t-value with 10 d.o.f. = 3.169

Since the test statistic does not exceed the critical t-value, we fail to reject Ho.

Decision: Null hypothesis accepted.


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9. Calorie Watchers Summary of the data:

Name Weight Change

Morco GainedLim Lost

Tan No changeYap GainedCruz LostMoran GainedChan GainedCruz LostAn LostChan Lost

Null: There is no significant relationship with the weight before and after the three week period

Alternative: there is a significant relationship with the weight before and after the three week period

10(313)-(65)(65)

R= --------------------------------

10(523)(65) 10(396)(58)

= -0.827

Ho: p.>0; H1: P< 0. Reject Ho if t< -2.896

-0827 10-2

t = ----------------------------- = -4.16

1- (-0.827)

Decision: Reject Ho. There is a negative association between months owned and hours exercised.

There is an inverse relationship between the variables. As the months increase, the numbers of

exercise decreases.


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10.Quality control department town problemNational News Sports

Undercharge 20 10Overcharge 15 30

Correct Price 200 225Total 235 265

Please refer to No. 2

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