Statistics

Statistics 3

Question 1:

Subject

Mean (u)

Standard deviation (o)

Psychology

Statistics

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Assume normal distribution, then

68% of scores within I SD

95% of scores within 2 SD

99.7% of scores within3 SD

Susan obtained 75 in Psychology and 70 in statistics,

a. In which subject did she do better?

In psychology 99.7% of the scores are between 65 + 3 * 10 = 95 and 65 – 3 *10 = 35

In statistics 99.7% of the scores are between 60 + 3 * 5= 75 and 60 – 3 *5 = 45

99.7% of the score in

Psychology range from 35 to 95

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Statistics range from 45 to 75

From the above it is evident that the upper limit of psychology scores is 95% while the upper limit of statistics is approximately 75, for this reason therefore Susan performed better in statistics than psychology.

Answer: statistics

b. What percentage of students did she perform better than in Psychology?

Using the normal distribution values can be standardized and probabilities obtained, values are standardized using the formula:

Z = X – mean/ standard deviation

In this case x is the score obtained in psychology = 70, therefore

Z = 75 – psychology score mean/ psychology score standard deviation

Z = 75 – 65/10

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Z=10/10=1

Using the Z table the probability of 1 = 0.3413(34.13%), this is the probability for the score between the mean value and the score (75), the probability below the mean is 50%, therefore Susan outperformed 50 + 34.13 = 84.13%.

Answer 69.15%

c. What percentage of students performed better than Susan in Statistics?

Z = X – mean/ standard deviation

In this case x is the score obtained in statistics = 70, therefore

Z = 70 – statistics score mean/ statistics score standard deviation

Z=70–60/5

Z=10/5=2

Using the Z table the probability of 1 = 0.4772(47.72%), this is the probability for the score between the mean value and the score (70), the probability below the mean is 50%; therefore Susan outperformed 50 + 47.72 = 97.72%. those who performed better than her were 100 – 97.72%. = 2.28%,

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Answer 2.28%

Question 2:

N= 1200

Mean = 175 cm

Standard deviation = 5 cm.

(a)The proportion of Navy Officers with heights between 170 cm and 180 cm

The following is the distribution curve:

Z = X – mean/ standard deviation

170:

170-175/5= -1

180:

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180-175/5= 1

From Z table the value 1 score is0.3413

For both -1 and 1 the value is 0.3433

From the distribution above the area is derived by adding up both values, therefore the proportion between 170 and 180 = 0.3433+0.3433 = 0.6466 or 64.66%

Answer =64.66%

(b) Percentage with heights greater than 183.5 cm

Height greater than 183.5

Z = X – mean/ standard deviation

= 183.5 -170/ 5 = 13.5/5 = 2.7

From Z table the value 2.7 area is = 0.4965

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Area below 183.5 = 0.5 + 0.4965 = 0.9965

Area above 183.5 = 1 – 0.9965 = 0.0035

Percentage of officers whose height is greater than 183.5 cm = 0.35%

Answer =0.35%

Z = X – mean/ standard deviation

= 185 -170/ 5 = 15/5 = 3

From Z table the value 3 area is = 0.4987

Area below 183.5 = 0.5 + 0.4987 = 0.9987

Area above 183.5 = 1 – 0.9987 = 0.0013

Percentage of officers whose height is greater than 185cm = 0.13%

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Total n = 1200

Number of officers whose height is greater than 185 cm = 0.13 * 1200 = 1.56

Rounding off to a whole number the number is 2

Answer = 1.56, rounding off gives us 2 officers

Question 3:

For a normal population with a mean u = 100 and a standard deviation o = 12:

(a) What is the probability of obtaining a sample mean greater than 106 for a sample of n = 4 scores?

Probabilities of obtaining mean greater than 106:

Z = X – mean/ standard deviation

= 106 -100/ 12 = 6/12 = 0.5 From z table the value 0.5

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Z = 0.1915

Greater than 106

= 0.5 + 0.1915 = 0.6519 Probability = 65.19%

(b) What is the probability of obtaining a sample mean less than 106 for a sample of n = 16 scores?

Probabilities of obtaining mean less than 106 = 1 – probability of obtaining greater than 106

=1-0.6519 = 0.3481

Probability = 34.81%

Question 4:

Seven sixth-grade children were randomly selected to participate in a teaching experiment. At the end of the experiment, the children completed a test to assess sixth-grade school achievement which has been standardized to have a mean u = 300.

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The resulting achievement test scores are:

386	311	374	316	303
298	387

Showing all workings, determine whether the data provides sufficient evidence of a change in the average test scores of the participants? Test at the 5% level of significance using a two-tailed test.

Total score for the sample = 386 + 311 + 374 + 316 + 303 + 298 + 387 = 2375

Mean of the sample = 2375/7 = 339.2857

a. Hypotheses:

H0: M1 = M2

H1: M1 ≠ M2

Where M1 is population mean and M2 is the sample mean.

b. Level of significance:

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5% level of significance two tails

c. Test statistics:

Total score for the sample = 386 + 311 + 374 + 316 + 303 + 298 + 387 = 2375

Mean of the sample = 2375/7 = 339.2857

Mean (M) = 339.2857

Number of participants (n) = 7

Estimated standard error (s M) =

Standard error = standard deviation / square root sample size

Standard deviation = square root (sum(x – mean) 2)/n)

x-mean

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x-mean2

x-mean2/n

386

46.71429

2182.224

311.7464

311

-28.2857

800.0816

114.2974

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374

34.71429

1205.082

172.1545

316

-23.2857

542.2245

77.46064

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303

-36.2857

1316.653

188.0933

298

-41.2857

1704.51

243.5015

387

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47.71429

2276.653

325.2362

1432.49

37.84825

Standard deviation = square root (1432.49)

Standard deviation =37.84825

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Standard error = standard deviation / square root sample size

Standard error = 37.84825/ square root sample (7)

Standard error = 14.30529d. Degrees of freedom (df) = 7-1=6

e. T statistic

T statistic = M2 – M1/ standard error

T statistic = 339.2857 – 300/ 14.30529

T statistic = 2.746236

f. Critical value:

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From the T table level of significance at 5% two tail, 6 degrees of freedom (t critical)

Critical value = 2.44691

g. Decision and conclusion:

T statistics is greater than the critical value, the decision is to reject the null hypothesis, and by rejecting null hypothesis this means that the sample mean value is not equal to the estimated population mean at the 5% level of significance.

Question 5:

Participant

Cigarettes consumed before treatment

After treatment

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sum

209

172

mean

23.22222

19.11111

standard deviation

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6.68539

7.720823

a. Hypotheses:

H0: M1 = M2

H1: M1 ≠ M2

Where M1 is the number of cigarettes smoked before treatment and M2 is the number of cigarettes smoked after treatment.

b.Level of significance:

a = 1% level of significance

c. Test statistics:

Mean (M)

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M1 =23.22222

M2 = 19.11111

Difference = 4.1111

d. Number of participants (n) = 9

e. Estimated standard error (s M) =

Standard error = square root [(variance 1 / sample size1) + (variance 2 / sample size2)]

Standard error = square root [(44.69444/ 9) + (59.61111/ 9)]

Standard error = square root [(4.966049) + (6.623457)]

Standard error = square root [11.58951]

Standard error = 3.404336

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f. Degrees of freedom (df) = 9-1 = 8

t = statistic =

t statistics = mean difference/ standard error

t statistics = 4.1111/ 3.404336

t statistics = 1.20761

g. Critical value:

From the T table level of significance at 1% two tail, 8 degrees of freedom (t critical)

Critical value =3.3553

h. Decision and conclusion:

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T critical value is greater than the T statistics value, for this reason the null hypothesis is accepted, therefore at the 1% level of significance the two mean values are equal and therefore hypnotherapy has no significant impact at the 1% level of significance.

Question 6:

Scores on a reading comprehension test were recorded for two groups of nine students enrolled in a remedial reading class. One group read each test passage aloud before answering the comprehension questions, the other read silently. The resulting test scores are given in Table 1 below.

Reading test scores

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Aloud

Silent

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Does the data support the researcher’s contention that the mean comprehension score will differ for the two reading methods. Test at the 5% level of significance using a two-tailed test.

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a. Hypotheses: (symbol format)

H0: M1 = M2

H1: M1 ≠ M2

Where M1 is the mean score for those who read the passage aloud and M2 is the mean score for those who read the passage silently

b. Level of significance:

a = 5%

c. Test statistic:

Reading test scores

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Aloud

Silent

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sum

283

316

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mean

31.44444

35.11111

standard deviation

4.558265

5.060742

variance

20.77778

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25.61111

M1 = 31.44444

SS1 = 4.558265 df1 = 9 DF =9-1=8

M2 = 35.11111 SS2 = 5.060742 df2 = 9

sp2=

Standard error = square root [(variance 1 / sample size1) + (variance 2 / sample size2)]

Standard error = square root [(20.77778/ 9) + (25.61111/ 9)]

Standard error = square root [(2.308642) + (2.845679)]

Standard error = square root [5.154321]

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Standard error = 2.270313

d. S (M1-M2) = 31.44444-35.11111=-3.66667

e. t statistics

T statistics = mean difference/ standard error

T statistics = 3.66667/ 2.270313

T statistics = -1.61505

f. Critical value:

From the T table level of significance at 5% two tail, 8 degrees of freedom (t critical)

T critical =2.3060

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g. Decision and conclusion:

From the above the t statistics value-1.61505 and lies in the acceptance region, therefore the null hypothesis that the two mean values are equal is accepted, this means that there is difference in score whether a passage is read silently or aloud at the 5% level of significance.

Question 7:

The Statistics marks (out of 10) and Psychology grades (out of 5) of eight students are given in Table 1 below.

Table 1: Statistics and Psychology Scores for Eight Students

Statistics (X)

Psychology (Y)

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(a) Construct a scatter plot using the data in Table 1. Include appropriate labels on both axes.

The scatter diagram below was produced using excel:

(b) Is there a linear relationship between the two sets of scores?

There is a linear relationship between the two scores, from the chart the scatter plot depict that as one score increases the other score also increases.

Correlation coefficient = square root [covariance (XY)] 2/ [(covariance xx)* (covariance yy)]

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Covariance x = (sumx2) – (nx’2)

Covariance y = (sum y2) – (ny’2)

Covariance xy= (sum xy) – nx’y’)

The table below summarises the results:

Statistics (X)

Psychology (Y)

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Statistics 3

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Statistics 3

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sum

278

100

164

mean

5.5

3.25

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standard deviation

2.267786838

1.488047618

Covariance x:

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Covariance x = (sumx2) – (nx’2)

Covariance x = (278) – (8*5.52)

Covariance x = (278) – (242)

Covariance x =36

Covariance y:

Covariance y = (sum y2) – (ny’2)

Covariance y = (100) – (8*3.252)

Covariance y = (100) – (84.5)

Covariance y = 15.5

Covariance xy:

Covariance xy= (sum xy) – (nx’y’)

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Covariance xy= (164) – (8*5.5*3.25)

Covariance xy= (164) – (143)

Covariance xy= 21

Correlation coefficient:

Correlation coefficient = square root [covariance (XY)] 2/ [(covariance xx)* (covariance yy)]

Correlation coefficient = square root [21 2/ [36*15.5]

Correlation coefficient = square root [0.790323]

Correlation coefficient = 0.889

(d) What is the critical level of Pearson’s r for this data? Use the .05 probability level.

DF=n-2

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DF=8-2=6

Using the moment correlation coefficient table the critical value is 0.707

(e) Is the result you calculated significant at the 5% level?

H0: r = 0

H1: r ≠ 0

Critical =0.707

When the correlation coefficient is greater than the critical value then the null hypothesis is rejected, in our case 0.889 > 0.707, therefore the coefficient is significant.

(f) What percentage of variation in Statistics scores?

Percentage of variation = r2

R2 = 0.889*0.889=0.790323

Percentage of variation = 79.03%

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References:

Gravetter and Forzano(2009) Research Methods for the Behavioral Sciences, New York:

McGraw hill

Gravetter and Wallnau (2009) Essentials of Statistics for the Behavioral Sciences, New York:

McGraw hill

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