Standard Deviation Reaction Times
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Question 1
A researcher is comparing the reaction times of 10yearolds who play video games and 10yearolds who do not. Random samples from each group provided the results shown below.
Reaction times of 10yearolds  
Players of
Video Games 
Nonplayers
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 nonplayers 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.
 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.
 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 crossclassified 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 hours^{2}.
 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.
 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 10yearolds who play video games and 10yearolds who do not. Random samples from each group provided the results shown below.
Reaction times of 10yearolds  
Players of
Video Games 
Nonplayers
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.
 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 10yearolds  
Players of
Video Games 
Nonplayers
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*(1α) % 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: = t_{0.01} = 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.
 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 nonplayers 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 nonplayers of video games can be stated as follows:
Null hypothesis, H_{0}:=
Alternate hypothesis, H_{1}: < (lefttailed or onetailed 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 (onetailed test): t_{0.025/2, 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 nonplayers 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.
 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*(1α) % 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 Z_{0.025/2} = Z_{0.0125} = 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.
 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, H_{0}: P_{1} = P_{2}
Alternate hypothesis, H_{1}: P_{1} ≠ P_{2} (twotailed test)
Level of significance: α = 1%
Test Statistic: = 3.65351 ≈ 3.65
Critical value of Z at 1% level: Z_{0.01/2} = ± 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, H_{0}: µ_{d }= 0
Alternative hypothesis, H_{a}: µ_{d}> 0 (Righttailed 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 tdistribution 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 ttest 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, H_{0}: The observed data do not contradict the hypothesized distribution of seasonal colds
Alternate hypothesis, H_{1}: The observed data contradicts the hypothesized distribution of seasonal colds
Chisquare statistic:
=, where O_{i} = observed frequency and E_{i} = Expected frequency
Calculation:
O  E  OE  (OE)^2  (OE)^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 Chisquare: At α = 0.05 level of significance, the critical value of Chisquare for 3 degrees of freedom is given by
= 7.815 (By referring Chisquare distribution table)
Decision Rule: Reject H_{0}, if the computed value of Chisquare statistic is greater than the critical value of 7.815.
Here, = 10.095 > 7.815
Conclusion:
Since the computed value of Chisquare 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 crossclassified 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, H_{0}: There is no relationship between an employee’s age and an employee’s opinion about the new wage incentive plan
Alternate hypothesis, H_{a}: 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: Chisquare test for independence of attributes
Chisquare statistic:
=, where O_{i} = observed frequency and E_{i} = 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
E_{i,j} = (ith row total * jth column total)/grand total
For example:
E_{1,1} = (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 Chisquare statistic:
O  E  OE  (OE)^2  (OE)^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 = (31)*(31) = 4 df
Critical Value of Chisquare: At α = 0.01 level of significance, the critical value of Chisquare for 4 degrees of freedom is given by
= 13.277 (By referring Chisquare distribution table)
Decision Rule: Reject H_{0}, if the computed value of Chisquare statistic is greater than the critical value of 13.277.
Here, = 17.352 > 13.277
Conclusion:
Since the computed value of Chisquare 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 hours^{2}.
 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).
 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.
H_{0}: σ = 10
H_{1}: σ < 10 (lefttailed test)
Test Statistic: = = 3.75
Critical value of Chisquare at 5% level:
= 7.26 (By referring Chisquare distribution table)
Decision Rule: Reject H_{0}, if the computed value of Chisquare statistic is less than the critical value of 7.26.
Here, = 3.75 < 7.26 è reject H_{0}
Conclusion:
Since the computed value of Chisquare 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, H_{0}: All mean times to relieve pain are equal
Alternate hypothesis, H_{1}: Not all mean times to relieve pain are equal
Test to be applied: Oneway 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*(12.7517.33)^{2} + 4*(20.2517.33)^{2} + 4*(2917.33)^{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  Pvalue  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: F_{2.5%, (2,9)df} = 5.715 (by referring Ftable)
Here, F = 3.3892 < 5.715 è fail to reject H_{0}
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.