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Chemical Reaction Experiment: Effect of Substrate Concentration

A chemist suspects the concentration of a substrate will have an effect on the velocity of a chemical reaction. To test her suspicion, she runs the reaction 15 times using a substrate concentration of 1.5 moles per liter and then 11 times using a substrate concentration of 2.0 moles per liter, each time measuring reaction velocity. This is thought to generate independent random samples from normal populations.

The data (in micromoles per 30 minutes) are given below:

Substrate Concentration: 1.5 moles per liter	Substrate Concentration: 2.0 moles per liter
6.7	7.5
7.1	9.1
7.9	11.7
6.7	8.9
6.4	6.1
9.0	8.1
4.8	9.1
5.0	5.7
6.8	8.3
9.6	2.0
8.9	11.0
6.0
6.7
7.6
7.3

Summary Statistics

• For substrate concentration 1.5 moles per liter:

  • Mean: ( \bar{y}_{1.5} = 7.10 )
  • Standard deviation: ( s_{1.5} = 1.3628 )

• For substrate concentration 2.0 moles per liter:

  • Mean: ( \bar{y}_{2.0} = 8.41 )
  • Standard deviation: ( s_{2.0} = 1.8322 )

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Raw Data

Velocity	Concentration
6.7	1.5
7.1	1.5
7.9	1.5
6.7	1.5
6.4	1.5
9.0	1.5
4.8	1.5
5.0	1.5
6.8	1.5
9.6	1.5
8.9	1.5
6.0	1.5
6.7	1.5
7.6	1.5
7.3	1.5
7.5	2.0
8.9	2.0
6.1	2.0
8.1	2.0
9.1	2.0
5.7	2.0
8.3	2.0
11.0	2.0
11.7	2.0
8.9	2.0
7.2	2.0

Hypothesis Testing for Substrate Concentration and Reaction Velocity

Using the two-sample pooled t-test, test whether or not the substrate concentration has an effect on reaction velocity. The significance level for the test is ( \alpha = 0.05 ). Choose the correct alternative hypotheses:

• $H_a:{\mu }_{1.5}<{\mu }_{2.0}$
• $H_a:{\mu }_{1.5}>{\mu }_{2.0}$
• $H_a:{\mu }_{1.5}\ne {\mu }_{2.0}$
• $H_a:{\mu }_{1.5}={\mu }_{2.0}$

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Test Statistic Calculation

Calculate the test statistic either using software or by hand for the test of whether or not the substrate concentration has an effect on reaction velocity.
Round your answer to two decimal places.

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P-Value Selection

Select the appropriate p-value for the hypothesis test that tests if the two mean reaction velocities differ:

• 0.0471
• < 0.0001
• 0.0671
• 0.2874

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Hypothesis Test Conclusion

Based on the p-value chosen in Question 3, select the appropriate conclusion for the hypothesis test.
$\alpha =0.05$

• Fail to reject $H_0$. Conclude that the true mean reaction velocities are equal between concentrations of 1.5 and 2.0 moles per liter.
• Fail to reject $H_0$. Conclude that there is not enough evidence to suggest a difference between the two mean reaction velocities for the two concentrations.
• Reject $H_0$. Conclude that the true mean reaction velocities differ between concentrations of 1.5 and 2.0 moles per liter.
• Reject $H_0$. Conclude that the true mean reaction velocities are the same between concentrations of 1.5 and 2.0 moles per liter.

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Confidence Interval Calculation

Construct the 95% confidence interval for the difference in mean reaction velocity. What is the lower bound for this interval?
(Round answer to 2 decimal places.)

$$\text{Answer to Question 5}\le \mu \le \text{Answer to Question 6}$$

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Upper Bound of Confidence Interval

What is the upper bound for this interval?
(Round answer to 2 decimal places.)

$$\text{Answer to Question 5}\le \mu \le \text{Answer to Question 6}$$

Consulting Problem: Sample Size Calculation

You are working as a consulting statistician for an educational research group. The group plans to implement a new teaching strategy and compare that strategy with the current teaching method. The metric used for the comparison will be test score.

The team is interested in seeing a difference in test scores of 18 points between the two independent groups. Historical records indicate that the standard deviation for test scores is 10 points.

Use this problem to answer Question 7.

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Sample Size Calculation

What sample size will be needed for each group for a power of 0.90 and ( \alpha = 0.05 )?
(Remember, if using JMP, the software will give you the total sample size needed, so you will need to divide that number by 2 to get the sample size in each group.)

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Ans:

Answer to Question 1:

To test whether the substrate concentration has an effect on reaction velocity, we need to choose the appropriate alternative hypothesis. Since we are testing if there is any difference (either increase or decrease) in mean reaction velocities between the two concentrations, the correct alternative hypothesis is:

• [X] $H_a:{\mu }_{1.5}\ne {\mu }_{2.0}$

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Answer to Question 2:

To calculate the test statistic for the two-sample pooled t-test, we’ll use the formula:

$$t=\frac{{\bar{y}}_1-{\bar{y}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$

Where:

• ${\bar{y}}_1=7.10$, $n_1=15$, $s_1=1.3628$
• ${\bar{y}}_2=8.41$, $n_2=11$, $s_2=1.8322$

Step 1: Calculate the pooled standard deviation ($s_p$)

$$s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}$$

Compute KaTeX parse error: Can't use function '\)' in math mode at position 7: s_1^2 \̲)̲ and \( s_2^2:

$$s_1^2=(1.3628{)}^2=1.856$$
$$s_2^2=(1.8322{)}^2=3.357$$

Calculate the numerator:

$$(15-1)(1.856)+(11-1)(3.357)=14\times 1.856+10\times 3.357=25.984+33.57=59.554$$

Calculate the denominator:

$$n_1+n_2-2=15+11-2=24$$

Compute $s_p$:

$$s_p=\sqrt{\frac{59.554}{24}}=\sqrt{2.481}=1.575$$

Step 2: Calculate the test statistic ($t$)

$$t=\frac{7.10-8.41}{1.575\times \sqrt{\frac{1}{15}+\frac{1}{11}}}$$

Calculate the standard error:

$$\sqrt{\frac{1}{15}+\frac{1}{11}}=\sqrt{\frac{11+15}{165}}=\sqrt{\frac{26}{165}}=\sqrt{0.1576}=0.397$$

Compute the denominator:

$$1.575\times 0.397=0.625$$

Compute $t$:
$$t=\frac{-1.31}{0.625}=-2.10$$

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Answer to Question 2: $-2.10$

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Answer to Question 3:

With the calculated test statistic $t=-2.10$ and degrees of freedom $df=24$, we can find the p-value.

Using a t-distribution table or calculator:

• P-value ≈ 0.0471

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Answer to Question 3: 0.0471

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Answer to Question 4:

Since the p-value $0.0471$ is less than the significance level $\alpha =0.05$, we reject the null hypothesis.

• Reject $H_0$. Conclude that the true mean reaction velocities differ between concentrations of 1.5 and 2.0 moles per liter.

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Answer to Question 4:

• Reject $H_0$. Conclude that the true mean reaction velocities differ between concentrations of 1.5 and 2.0 moles per liter.

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Answer to Question 5:

To construct the 95% confidence interval for the difference in mean reaction velocities, we use:

$$({\bar{y}}_1-{\bar{y}}_2)\pm t_{\alpha /2,\nu }\times s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$$

Where:

• $t_{\alpha /2,\nu }$ is the critical t-value for $\alpha /2=0.025$ and $\nu =24$ degrees of freedom. From t-tables, $t_{0.025,24}=2.064$.
• We already have $s_p=1.575$ and the standard error $SE=0.625$.

Calculate the margin of error (ME):

$$ME=t_{0.025,24}\times SE=2.064\times 0.625=1.29$$

Compute the confidence interval:

$$\text{Lower bound}=({\bar{y}}_1-{\bar{y}}_2)-ME=-1.31-1.29=-2.60$$

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Answer to Question 5: $-2.60$

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Answer to Question 6:

$$\text{Upper bound}=({\bar{y}}_1-{\bar{y}}_2)+ME=-1.31+1.29=-0.02$$

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Answer to Question 6: $-0.02$

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Final Confidence Interval:

$$-2.60\le {\mu }_{1.5}-{\mu }_{2.0}\le -0.02$$

This interval does not include zero, which aligns with our earlier conclusion to reject the null hypothesis.

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Apologies for the confusion earlier.

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Answer to Question 7:

To determine the required sample size for each group, we’ll use the formula for the sample size needed to detect a difference between two independent means with a specified power and significance level:

$$n={\left(\frac{(Z_{1-\alpha /2}+Z_{1-\beta })\times \sigma \sqrt{2}}{\Delta }\right)}^2$$

Given:

• Desired difference to detect ($\Delta$): 18 points
• Standard deviation ($\sigma$): 10 points
• Significance level ($\alpha$): 0.05
• Power ($1-\beta$): 0.90

Step 1: Find the Z-scores

• For $\alpha =0.05$ (two-tailed test):

  $$Z_{1-\alpha /2}=Z_{0.975}=1.96$$

• For power ( 1 - \beta = 0.90 ):

  $$Z_{1-\beta }=Z_{0.90}=1.28$$

Step 2: Calculate the sample size per group

Include the (\sqrt{2}) factor because we’re comparing two independent groups:

$$n={\left(\frac{(1.96+1.28)\times 10\times \sqrt{2}}{18}\right)}^2$$
Compute the numerator:

$$(1.96+1.28)\times 10\times \sqrt{2}=(3.24)\times 10\times 1.4142=3.24\times 14.142=45.85$$

Compute the fraction:

$$\frac{45.85}{18}=2.547$$

Square the result:

$$(2.547{)}^2=6.484$$

Therefore, the required sample size per group is approximately 6.484.

Since you cannot have a fraction of a participant, you must round up to the next whole number:

$$n=7$$

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Answer to Question 7:

• 7

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Each group will need 7 participants to achieve a power of 0.90 with a significance level of ( \alpha = 0.05 ) for detecting an 18-point difference in test scores.

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Explanation:

• Why include $\sqrt{2}$?

  When comparing two independent means, the standard error of the difference between means incorporates the variance from both groups. Therefore, the combined standard deviation for the difference is:

  $${\sigma }_{\text{diff}}=\sigma \times \sqrt{\frac{1}{n}+\frac{1}{n}}=\sigma \times \sqrt{\frac{2}{n}}=\frac{\sigma \sqrt{2}}{\sqrt{n}}$$

  This adjustment ensures that both groups’ variances are accounted for in the sample size calculation.

• Total Sample Size:

  The total sample size required is $N=2n=14$.

• Software Note:

  If using software like JMP, which provides the total sample size, you would get $N=14$, and dividing by 2 gives you $n=7$ per group.