Standard Deviation Reaction Times

Standard Deviation Reaction Times

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Question 1

A researcher is comparing the reaction times of 10-year-olds who play video games and 10-year-olds who do not. Random samples from each group provided the results shown below.

Reaction times of 10-year-olds
Players of

Video Games

Non-players

of Video Games

Sample size 15 10
Mean reaction time
(seconds)
       0.50        0.52
Standard deviation of reaction times (seconds)        0.01        0.02

 

Assume that the reaction times for both populations are normally distributed with the same population standard deviation.

a.Construct a 98% confidence interval for the difference between the population means of reaction times.

                           Note: Express your answer to 5 decimal places of accuracy.

b.Using a 2.5% significance level, can the researcher conclude that the mean reaction time of players of video games is less than the mean reaction time of non-players of video games? Formulate and test the appropriate hypotheses. Use the critical value approach.

Question 2

Is there any difference between the proportion of Americans who wish they were rich and the proportion of Canadians who wish they were rich? A random sample of 1,000 Americans showed that 550 wish they were rich, while a random sample of 500 Canadians showed that 225 wished they were rich.

  1. Find a 97.5% confidence interval for the difference between the population proportions who wish they were rich.

Note: Express your answer to 4 decimal places of accuracy.

  1. Test, at the 1% level of significance, whether the two population proportions are significantly different. Formulate and test the appropriate hypotheses. Use the critical value approach.

Question 3

A motivational program was designed to increase the number of hours each student who took it studied per week. Eight randomly selected students were involved in the program. The table below shows the number of hours each student studied per week before the taking motivational program and after taking the program.

 
Before program 8 12 10 12 5 10 10 12
After program 12 11 14 15 8 15 9 6

 

At the 10% level of significance, can it be concluded the motivational program increased the average number of hours students studied per week? Formulate and test the appropriate hypotheses. Use the critical value approach.

Assume that the population of paired differences has a normal distribution.

Question 4

It has been hypothesized that the distribution of seasonal colds in Canada is as follows:

Season Percentage
Fall 35%
Winter 25%
Spring 30%
Summer 10%

A random sample of 200 Canadian citizens provided the following results:

Season Observed Frequency
Fall 80
Winter 40
Spring 70
Summer 10

 

Do the observed data contradict the hypothesis? Formulate and test the appropriate hypotheses at the 5% level of significance. Use the critical value approach.

Question 5

A large company is interested in knowing whether there is any relationship between a person’s age and that person’s opinion about a new wage incentive plan. A random sample of 100 employees was selected and cross-classified as shown below.

    Opinion  
Opposed Neutral In Favour
Under 35   5   5 10
35 to 55 10 30 10
Over 55 15 10   5

 

At the 1% level of significance, can it be concluded that there is a relationship between an employee’s age and an employee’s opinion about the new wage incentive plan? Formulate and test the appropriate hypotheses. Use the critical value approach.

Question 6

A random sample of 16 students showed that the variance in the number of hours they spend studying for a final exam was 25 hours2.

  1. Construct a 95% confidence interval for the population variance and standard deviation of hours spent studying. Assume the population of hours spent studying is normally distributed.
  2. Test, at the 5% level of significance, whether the population standard deviation of hours spent studying is less than 10. Use the critical value approach.

Question 7

Three different brands of pain relief medication are available for minor aches and pains. It is of interest to know whether the mean time to relieve the pain differs from brand to brand. To find out, twelve individuals with minor aches and pains were randomly assigned to one of the three brands. The time to relieve the pain (in minutes) for each individual is provided in the table below.

 

Brand A

Brand B Brand C
10 15 20
12 22 25
15 25 20
14 19 11

Given that the necessary assumptions are satisfied, can it be concluded, at the 2.5% level of significance, that not all mean times to relieve pain are equal? Formulate and test the appropriate hypotheses. Use the critical value approach. 

Solution

Question 1

A researcher is comparing the reaction times of 10-year-olds who play video games and 10-year-olds who do not. Random samples from each group provided the results shown below.

  Reaction times of 10-year-olds
  Players of

Video Games

Non-players

of Video Games

Sample size 15 10
Mean reaction time
(seconds)
       0.50        0.52
Standard deviation of reaction times (seconds)        0.01        0.02

 

Assume that the reaction times for both populations are normally distributed with the same population standard deviation.

  1. Construct a 98% confidence interval for the difference between the population means of reaction times.

Solution:

The information given in the above problem can be represented with the standard notations as follows:

  Reaction times of 10-year-olds
  Players of

Video Games

Non-players

of Video Games

Sample size n1 = 15 n2 = 10
Mean reaction time
(seconds)
 = 0.50  = 0.52
Standard deviation of reaction times (seconds)  = 0.01  = 0.02

The 100*(1-α) % confidence interval for the difference between the population means is given by

where  and S =

The 98% confidence interval for the difference between the population means of reaction times is calculated as follows:

S =  = 0.014744

= 0.006019

Critical value of t at 2% level of significance:  = t0.01 = 2.500

By substituting appropriate values in the above confide3nce interval formula, we have

=

= (-0.02 – 0.015047,  -0.02 + 0.015047)

= (–0.03505, –0.00495)

Thus, the 98% confidence interval for the difference between the population means of reaction times is (–0.03505, –0.00495). This confidence interval does not contain zero, which indicates that the difference between the population means is statistically significant.

  1. Using a 2.5% significance level, can the researcher conclude that the mean reaction time of players of video games is less than the mean reaction time of non-players of video games? Formulate and test the appropriate hypotheses. Use the critical value approach.

Solution:

The hypotheses to test whether the mean reaction time of players of video games is less than the mean reaction time of non-players of video games can be stated as follows:

Null hypothesis, H0:=

Alternate hypothesis, H1: <       (left-tailed or one-tailed test)

Level of significance: α = 2.5%

Test Statistic:   = –3.32265 ≈ –3.32

Degrees of freedom: df = n1 + n2 -2 = 15 + 10 – 2 = 23

Critical value of t at 2.5% level (one-tailed test): t0.025/2, 23df = –2.0687

Rejection or critical region: {x: t < –2.0687}

Here, t = –3.32 < –2.0687 è Value of test statistic lies in the rejection region.

Conclusion: Since the computed value of the test statistic lies in the critical region, we reject the null hypothesis at 2.5% level of significance. Hence, the evidence is sufficient to conclude that the mean reaction time of players of video games is less than the mean reaction time of non-players of video games.

Question 2

Is there any difference between the proportion of Americans who wish they were rich and the proportion of Canadians who wish they were rich? A random sample of 1,000 Americans showed that 550 wish they were rich, while a random sample of 500 Canadians showed that 225 wished they were rich.

  1. Find a 97.5% confidence interval for the difference between the population proportions who wish they were rich.

Solution:

The information given in the above problem can be represented with the standard notations as follows:

Americans Canadians
Sample size n1 = 1000 n2 = 500
Rich (X) X1 = 550 X2 = 225
Sample Proportion p1 = 0.55 p2 = 0.45

The 100*(1-α) % confidence interval for the difference between the population proportions is given by

where P =  = = 0.51667

Q = 1 – P = 1 – 0.51667 = 0.48333

Now, the 97.5 % confidence interval for the difference between the population proportions is given by

where Z0.025/2 = Z0.0125 = 2.2414

= 0.0273709

By substituting appropriate values in the above confide3nce interval formula, we have

=

= (0.10 – 2.2414*0.0273709, 0.10 + 2.2414*0.0273709)

= (0.0387, 0.1613)

Thus, the 97.5% confidence interval for the difference between the population proportions who wish they were rich is (0.0387, 0.1613). This confidence interval does not contain zero, which indicates that the difference between two population proportions is statistically significant.

  1. Test, at the 1% level of significance, whether the two population proportions are significantly different. Formulate and test the appropriate hypotheses. Use the critical value approach. 

Solution:

The hypotheses to test whether the two population proportions are significantly different can be stated as follows:

Null hypothesis, H0: P1 = P2

Alternate hypothesis, H1: P1 ≠ P2        (two-tailed test)

Level of significance: α = 1%

Test Statistic  = 3.65351 ≈ 3.65

Critical value of Z at 1% level: Z0.01/2 = ± 2.576

Critical Region: {x: Z < -2.576 or Z > 2.576}

Here, the Z = 3.65 > 2.576 è Z lies in the critical region. 

Conclusion:

Since the computed value of Z lies in the critical region, we reject the null hypothesis at 1% level of significance. Hence, the evidence is sufficient to conclude that there is significant difference between the proportion of Americans who wish they were rich and the proportion of Canadians who wish they were rich.

Question 3

A motivational program was designed to increase the number of hours each student who took it studied per week. Eight randomly selected students were involved in the program. The table below shows the number of hours each student studied per week before the taking motivational program and after taking the program.

 
Before program 8 12 10 12 5 10 10 12
After program 12 11 14 15 8 15 9 6

 

At the 10% level of significance, can it be concluded the motivational program increased the average number of hours students studied per week? Formulate and test the appropriate hypotheses. Use the critical value approach.

Assume that the population of paired differences has a normal distribution.

Solution:

In order to test whether the motivational program increased the average number of hours students studied per week, the hypotheses can be stated as follows:

Hypotheses:

Null hypothesis, H0: µd = 0

Alternative hypothesis, Ha: µd> 0       (Right-tailed test)

Level of Significance: α = 0.10

Test Statistic:

To test the above null hypothesis, we consider the following test statistic:

t = ,                 where  and  are computed using the following formulas:

= and= Standard deviation of the difference (d) =

n = sample size = 8

The  and are computed using the following table:

Before program After program d (d – dbar)^2
8 12 4 6.891
12 11 -1 5.641
10 14 4 6.891
12 15 3 2.641
5 8 3 2.641
10 15 5 13.141
10 9 -1 5.641
12 6 -6 54.391
sum 11 97.875
mean 1.375
sd 3.739
n = 8

From the above table, we have

= = 11/8 = 1.375

= Standard deviation of the difference (d) = = = 3.739

On substituting the respective values, we have

t == = 1.0401

Thus the value of test statistic is t = 1.0401

Degrees of freedom = n – 1 = 8 – 1= 7

Level of significance = α = 0.10

Critical value of t:

At α = 0.10 level of significance, the critical value of t for 7 degrees of freedom are given by

= 1.415              (By referring t-distribution table)

The rejection region is given by

C = {X: t > 1.415}

Here, t = 1.0401 < 1.415 èt does not lie in the critical region

Conclusion:

Since the computed value of t-test statistic (t = 1.0401) does not lie in the critical region, we fail to reject the null hypothesis at 5% level of significance. Hence, the evidence is not sufficient to conclude that the motivational program increased the average number of hours students studied per week.

Question 4

It has been hypothesized that the distribution of seasonal colds in Canada is as follows:

Season Percentage
Fall 35%
Winter 25%
Spring 30%
Summer 10%

A random sample of 200 Canadian citizens provided the following results:

Season Observed Frequency
Fall 80
Winter 40
Spring 70
Summer 10

Do the observed data contradict the hypothesis? Formulate and test the appropriate hypotheses at the 5% level of significance. Use the critical value approach.

Solution:

Observed frequencies: 80, 40, 70 ,10

Total frequency: N = 80+40+70+10 = 200

Expected frequencies:

Fall = 35%*200 = 70

Winter = 25%*200 = 50

Fall = 30%*200 = 60

Fall = 10%*200 = 20

Hypotheses:

Null hypothesis, H0: The observed data do not contradict the hypothesized distribution of seasonal colds

Alternate hypothesis, H1: The observed data contradicts the hypothesized distribution of seasonal colds

Chi-square statistic:

=, where Oi = observed frequency and Ei = Expected frequency

Calculation:

O E O-E (O-E)^2 (O-E)^2 / E
80 70 10 100 1.429
40 50 -10 100 2.000
70 60 10 100 1.667
10 20 -10 100 5.000
200 200     10.095

From the above table, we have

= 10.095

Thus, the value of the test statistic is  = 10.095

Level of Significance: α = 0.05

Degrees of freedom: v = n – 1 = 4 – 1 = 3 

Critical Value of Chi-square: At α = 0.05 level of significance, the critical value of Chi-square for 3 degrees of freedom is given by

= 7.815                (By referring Chi-square distribution table)

Decision Rule: Reject H0, if the computed value of Chi-square statistic is greater than the critical value of 7.815.

Here,  = 10.095 > 7.815

Conclusion:

Since the computed value of Chi-square lies in the rejection region (i.e., 10.095 > 7.815), we reject the null hypothesis at 5% level of significance. Hence, the evidence is sufficient to conclude that the observed data contradicts the hypothesized distribution of seasonal colds.

Question 5

A large company is interested in knowing whether there is any relationship between a person’s age and that person’s opinion about a new wage incentive plan. A random sample of 100 employees was selected and cross-classified as shown below.

    Opinion  
Opposed Neutral In Favour
Under 35   5   5 10
35 to 55 10 30 10
Over 55 15 10   5

At the 1% level of significance, can it be concluded that there is a relationship between an employee’s age and an employee’s opinion about the new wage incentive plan? Formulate and test the appropriate hypotheses. Use the critical value approach.

Solution:

In order to test whether there is a relationship between an employee’s age and an employee’s opinion about the new wage incentive plan, the hypotheses can be stated as follows:

Hypotheses:

Null hypothesis, H0: There is no relationship between an employee’s age and an employee’s opinion about the new wage incentive plan

Alternate hypothesis, Ha: There is significant relationship between an employee’s age and an employee’s opinion about the new wage incentive plan

Level of significance: α = 0.01

Test to be applied: Chi-square test for independence of attributes 

Chi-square statistic:

=, where Oi = observed frequency and Ei = Expected frequency

Observed Frequency Table:

    Opinion  
Opposed Neutral In Favour Total
Under 35 5 5 10 20
35 to 55 10 30 10 50
Over 55 15 10 5 30
Total 30 45 25 100

Expected Frequency Calculation:

Expected frequency corresponding to (i, j)th cell is

Ei,j = (ith row total * jth column total)/grand total

For example:

E1,1 = (20*30)/100 = 6

E2,1 = (50*30)/100 = 15

Similarly, the expected frequency has been calculated for all the cells and expected frequency table is presented below:

Expected Frequency Table:

    Opinion  
Opposed Neutral In Favour Total
Under 35 6 9 5 20
35 to 55 15 22.5 12.5 50
Over 55 9 13.5 7.5 30
Total 30 45 25 100

Calculation of Chi-square statistic:

O E O-E (O-E)^2 (O-E)^2 / E
5 6 -1 1 0.167
10 15 -5 25 1.667
15 9 6 36 4.000
5 9 -4 16 1.778
30 22.5 7.5 56.25 2.500
10 13.5 -3.5 12.25 0.907
10 5 5 25 5.000
10 12.5 -2.5 6.25 0.500
5 7.5 -2.5 6.25 0.833
100 100     17.352

From the above table, we have

= 17.352

Thus, the value of the test statistic is = 17.352

Level of Significance: α = 0.01

Degrees of freedom: v = (3-1)*(3-1) = 4 df 

Critical Value of Chi-square: At α = 0.01 level of significance, the critical value of Chi-square for 4 degrees of freedom is given by

= 13.277              (By referring Chi-square distribution table)

Decision Rule: Reject H0, if the computed value of Chi-square statistic is greater than the critical value of 13.277.

Here,  = 17.352 > 13.277

Conclusion:

Since the computed value of Chi-square lies in the rejection region (i.e., 17.352 > 13.277), we reject the null hypothesis at 5% level of significance. Hence, the evidence is sufficient to conclude that there is significant relationship between an employee’s age and an employee’s opinion about the new wage incentive plan.

Question 6

A random sample of 16 students showed that the variance in the number of hours they spend studying for a final exam was 25 hours2.

  1. Construct a 95% confidence interval for the population variance and standard deviation of hours spent studying. Assume the population of hours spent studying is normally distributed.

Solution:

The information given in the problem is represented with the standard notations as follows:

Sample size: n = 16 students

Sample standard deviation: s = 25 hours

To find a 95% confidence interval for the population variance and standard deviation of hours spent studying:

Formula for 95% confidence interval for the variance is given as follows:

=, where  = 27.488 and  = 6.262

=

= (13.642, 59.884)

Thus, the 95% confidence interval for the population variance of hours spent studying is (13.642, 59.884).

Formula for 95% confidence interval for the standard deviation is given as follows:

=

= (3.694, 7.738)

Thus, the 95% confidence interval for the population standard deviation of hours spent studying is (3.694, 7.738).

  1. Test, at the 5% level of significance, whether the population standard deviation of hours spent studying is less than 10. Use the critical value approach.

H0: σ = 10

H1: σ < 10       (left-tailed test)

Test Statistic:  =  = 3.75

Critical value of Chi-square at 5% level:

= 7.26               (By referring Chi-square distribution table)

Decision Rule: Reject H0, if the computed value of Chi-square statistic is less than the critical value of 7.26.

Here,  = 3.75 < 7.26 è reject H0 

Conclusion:

Since the computed value of Chi-square lies in the rejection region (i.e., 3.75 < 7.26), we reject the null hypothesis at 5% level of significance. Hence, the evidence is sufficient to conclude that the population standard deviation of hours spent studying is less than 10.

Question 7

Three different brands of pain relief medication are available for minor aches and pains. It is of interest to know whether the mean time to relieve the pain differs from brand to brand. To find out, twelve individuals with minor aches and pains were randomly assigned to one of the three brands. The time to relieve the pain (in minutes) for each individual is provided in the table below.

Brand A Brand B Brand C
10 15 20
12 22 25
15 25 20
14 19 11

Given that the necessary assumptions are satisfied, can it be concluded, at the 2.5% level of significance, that not all mean times to relieve pain are equal? Formulate and test the appropriate hypotheses. Use the critical value approach.

Solution:

Hypotheses:

Null hypothesis, H0: All mean times to relieve pain are equal

Alternate hypothesis, H1: Not all mean times to relieve pain are equal

Test to be applied: One-way ANOVA

Level of significance: α = 2.5%

S. No. X Y Z X^2 Y^2 Z^2
1 10 15 20 100 225 400
2 12 22 25 144 484 625
3 15 25 20 225 625 400
4 14 19 11 196 361 121
SUM 51 81 76 665 1695 1546
MEAN 12.75 20.25 19 166.25 423.75 386.5

Grand average = (12.75 + 20.25 + 19)/3 = 17.3333

SSTR = 4*(12.75-17.33)2 + 4*(20.25-17.33)2 + 4*(29-17.33)2 = 129.1667

SSE = 171.50              (SS for error)

SST = 300.667            (SS for total)

ANOVA Table:

Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Brand A 4 51 12.75 4.9167
Brand B 4 81 20.25 18.2500
Brand C 4 76 19 34
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 129.1667 2.0000 64.5833 3.3892 0.0799 4.2565
Within Groups 171.5000 9.0000 19.0556
Total 300.6667 11

From the above ANOVA table, we have

F = 3.3892

Critical value of F: F2.5%, (2,9)df = 5.715           (by referring F-table)

Here, F = 3.3892 < 5.715 è fail to reject H0

Conclusion:

Since the computed value of F does not lie in the critical region, we fail to reject the null hypothesis at 2.5% level of significance. Hence, the evidence is not sufficient to conclude that not all mean times to relieve pain are equal.