To calculate an independent sample t-test with equal variance, you will need to follow these steps:
Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha).
In this case, the null hypothesis would be that there is no difference in the mean scores between the CLOCKWISE and COUNTER-CLOCKWISE conditions. The alternative hypothesis would be that there is a difference in the mean scores between the two conditions.
H0: μ1 = μ2
Ha: μ1 ≠ μ2
Step 2: Calculate the mean and standard deviation for each sample.
For the CLOCKWISE condition:
Sample mean (x̄1) = 165.29
Sample standard deviation (s1) = 10.65
For the COUNTER-CLOCKWISE condition:
Sample mean (x̄2) = 148.67
Sample standard deviation (s2) = 2.81
Step 3: Calculate the pooled standard deviation (Sp).
Sp = sqrt(((n1-1)s1^2 + (n2-1)s2^2)/(n1 + n2 - 2))
= sqrt(((50-1)(10.65^2) + (60-1)(2.81^2))/(50 + 60 - 2))
= sqrt((49(113.4225) + 59(7.8961))/108)
= sqrt((5563.9175 + 465.2039)/108)
= sqrt(6029.1214/108)
= sqrt(55.8413)
≈ 7.47
Step 4: Calculate the t-statistic.
t = (x̄1 - x̄2) / (Sp * sqrt(1/n1 + 1/n2))
= (165.29 - 148.67) / (7.47 * sqrt(1/50 + 1/60))
= 16.62 / (7.47 * 0.3129)
= 16.62 / 2.3364
≈ 7.10
Step 5: Determine the critical t-value and degrees of freedom.
To determine the critical t-value, we need to know the significance level and the degrees of freedom. Let's assume a significance level of 0.05.
Degrees of freedom (df) = n1 + n2 - 2
= 50 + 60 - 2
= 108
Critical t-value (two-tailed) = ±2.000 (approximately), with 108 degrees of freedom for a significance level of 0.05.
Step 6: Compare the calculated t-value with the critical t-value.
Since the calculated t-value (7.10) is greater than the critical t-value (±2.000), we can reject the null hypothesis.
Step 7: Analyze the result.
Based on the results, there is sufficient evidence to conclude that there is a significant difference in the mean scores between the CLOCKWISE and COUNTER-CLOCKWISE conditions.
calculate independent sample t test for an equal variance with the following data set.
CLOCKWISE(X)
185
182
184.3
176
179.5
178.2
181.7
159.5
162.3
160.5
166
163.5
162.7
167
167
163
162.5
162
163
161
163.5
160.3
167.3
127.8
128.8
137
113.2
116.5
131.5
144.5
143.3
152
143
141.2
141
142.6
141.7
126.6
128
140
141.6
142
138
140
138.4
142
137.8
140.4
137.4
134.2
COUNTER-CLOCKWISE(Y)
136.8
138.8
139
138.8
136.8
152.2
152.4
153.2
152.8
151.6
149.6
152.8
152.6
149.2
150.6
150
148.6
149.4
149.6
149.6
151
149.2
148.2
152.2
150.2
150.8
149.4
148.2
150.2
148.6
149
150.2
148.6
150.8
149
149.8
146.6
150.2
148.4
149.2
149.2
148.8
150.2
148.4
148.4
149.6
148.8
147.2
148.8
154
150.4
148
144
149.2
148.2
149
149.6
146.8
148.6
149.8
149
148.2
147.8
147.2
147
149.8
148.2
146.4
146
147.2
148.1
146.1
150
146.8
152
149
148.8
149.2
147.9
151.2
146.6
149.7
148.3
147.5
148.7
147.9
148.7
149.7
147.5
151.6
150.9
150.4
150.7
148.5
146.7
4 answers
1.give the concentration of each of the following type of ion in the following solution .
a.0.70M Co(NO3)2
b.1.4M Fe(ClO4)3
c.2.75M Al2(SO4)3
2.In the reaction 3O2 to produce 2O3 the rate of appearance of ozone was 1.5mol/min .what is the rate of disappearance of oxygen.
3. calculate the PH of an aqueous solution consisting of 0.06M acetic acid and 0.08M sodium acetate.
a.0.70M Co(NO3)2
b.1.4M Fe(ClO4)3
c.2.75M Al2(SO4)3
2.In the reaction 3O2 to produce 2O3 the rate of appearance of ozone was 1.5mol/min .what is the rate of disappearance of oxygen.
3. calculate the PH of an aqueous solution consisting of 0.06M acetic acid and 0.08M sodium acetate.
1. To find the concentration of each ion in the given solutions, we need to consider the dissociation of the compounds.
a. Co(NO3)2 dissociates into Co2+ and 2 NO3- ions. Therefore, the concentration of Co2+ is 0.70 M, and the concentration of NO3- is 2 x 0.70 M = 1.40 M.
b. Fe(ClO4)3 dissociates into Fe3+ and 3 ClO4- ions. Therefore, the concentration of Fe3+ is 1.4 M, and the concentration of ClO4- is 3 x 1.4 M = 4.2 M.
c. Al2(SO4)3 dissociates into 2 Al3+ and 3 SO4^2- ions. Therefore, the concentration of Al3+ is 2 x 2.75 M = 5.50 M, and the concentration of SO4^2- is 3 x 2.75 M = 8.25 M.
2. The balanced equation for the reaction is: 3O2 -> 2O3.
From the equation, we can see that for every 3 moles of O2 that react, 2 moles of O3 are produced. Therefore, the rate of disappearance of oxygen is directly proportional to the rate of appearance of ozone, but with a different stoichiometric ratio.
Rate of disappearance of oxygen = (2/3) x 1.5 mol/min = 1 mol/min.
3. To calculate the pH of the given solution consisting of acetic acid (CH3COOH) and sodium acetate (CH3COONa), we need to consider the dissociation of acetic acid and the common ion effect.
Acetic acid (CH3COOH) dissociates as follows: CH3COOH -> CH3COO- + H+
Sodium acetate (CH3COONa) dissociates as follows: CH3COONa -> CH3COO- + Na+
Since sodium acetate dissociates completely, the concentration of CH3COO- ions is equal to the concentration of sodium acetate, which is 0.08 M.
Using the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid])
The pKa of acetic acid is approximately 4.76.
pH = 4.76 + log(0.08 M/0.06 M)
= 4.76 + log(1.33)
≈ 4.76 + 0.124
≈ 4.88
Therefore, the pH of the aqueous solution consisting of 0.06 M acetic acid and 0.08 M sodium acetate is approximately 4.88.
a. Co(NO3)2 dissociates into Co2+ and 2 NO3- ions. Therefore, the concentration of Co2+ is 0.70 M, and the concentration of NO3- is 2 x 0.70 M = 1.40 M.
b. Fe(ClO4)3 dissociates into Fe3+ and 3 ClO4- ions. Therefore, the concentration of Fe3+ is 1.4 M, and the concentration of ClO4- is 3 x 1.4 M = 4.2 M.
c. Al2(SO4)3 dissociates into 2 Al3+ and 3 SO4^2- ions. Therefore, the concentration of Al3+ is 2 x 2.75 M = 5.50 M, and the concentration of SO4^2- is 3 x 2.75 M = 8.25 M.
2. The balanced equation for the reaction is: 3O2 -> 2O3.
From the equation, we can see that for every 3 moles of O2 that react, 2 moles of O3 are produced. Therefore, the rate of disappearance of oxygen is directly proportional to the rate of appearance of ozone, but with a different stoichiometric ratio.
Rate of disappearance of oxygen = (2/3) x 1.5 mol/min = 1 mol/min.
3. To calculate the pH of the given solution consisting of acetic acid (CH3COOH) and sodium acetate (CH3COONa), we need to consider the dissociation of acetic acid and the common ion effect.
Acetic acid (CH3COOH) dissociates as follows: CH3COOH -> CH3COO- + H+
Sodium acetate (CH3COONa) dissociates as follows: CH3COONa -> CH3COO- + Na+
Since sodium acetate dissociates completely, the concentration of CH3COO- ions is equal to the concentration of sodium acetate, which is 0.08 M.
Using the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[acid])
The pKa of acetic acid is approximately 4.76.
pH = 4.76 + log(0.08 M/0.06 M)
= 4.76 + log(1.33)
≈ 4.76 + 0.124
≈ 4.88
Therefore, the pH of the aqueous solution consisting of 0.06 M acetic acid and 0.08 M sodium acetate is approximately 4.88.
4. Calculate the [𝐻^+ ] 𝑎𝑛𝑑 [𝑁𝑂_3^( −) ] of a 0.175M solution of nitric acid, 𝑁𝐻𝑂_3^( −).
5. Calculate the hydroxide ion concentration in a 0.175 solution of HCl.
6. Calculate the hydrogen ion concentration in 0.01M solution of Ca(𝑂𝐻)_2
7. The ionisation constant of water is 𝐾_𝑤=1.00×10^(−13.60) at body temperature, 37^0 𝐶. 𝑊ℎ𝑎𝑡 𝑎𝑟𝑒 𝑡ℎ𝑒 𝐻_3 𝑂^+ 𝑎𝑛〖 𝑂𝐻〗^− concentrations at that temperature?
8. Calculate the pH of the following solutions at 25^0 𝐶:(𝑎)0.0028𝑀 𝐻𝐶𝑙, (𝑏)0.014 𝐻𝑁𝑂_3, (𝑐)0.00052𝑀 𝐻𝐵𝑟,
0.092𝑀 𝐻𝐼
9. Calculate the pH of a 0.025M solution of ammonia, 𝐾_𝑏=1.75× 10^(−5)
5. Calculate the hydroxide ion concentration in a 0.175 solution of HCl.
6. Calculate the hydrogen ion concentration in 0.01M solution of Ca(𝑂𝐻)_2
7. The ionisation constant of water is 𝐾_𝑤=1.00×10^(−13.60) at body temperature, 37^0 𝐶. 𝑊ℎ𝑎𝑡 𝑎𝑟𝑒 𝑡ℎ𝑒 𝐻_3 𝑂^+ 𝑎𝑛〖 𝑂𝐻〗^− concentrations at that temperature?
8. Calculate the pH of the following solutions at 25^0 𝐶:(𝑎)0.0028𝑀 𝐻𝐶𝑙, (𝑏)0.014 𝐻𝑁𝑂_3, (𝑐)0.00052𝑀 𝐻𝐵𝑟,
0.092𝑀 𝐻𝐼
9. Calculate the pH of a 0.025M solution of ammonia, 𝐾_𝑏=1.75× 10^(−5)
1 answer