91Ó°ÊÓ

A study is made to see if increasing the substrate concentration has an appreciable effect on the velocity of a chemical reaction. With a substrate concentration of 1.5 moles per liter, the reaction was run 15 times with an average velocity of 7.5 micromoles per 30 minutes and a standard deviation of 1.5. With a substrate concentration of 2.0 moles per liter, 12 runs were made, yiclding an average velocity of 8.8 micromoles per 30 minutes and a sample standard deviation of 1.2 . Is there any reason to believe that this increase in substrate concentration causes an increase in the mean velocity by more than 0.5 micromole per 30 minutes? Use a 0.01 level of significance and assume the populations to be approximately normally distributed with equal variances.

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis, denoted \(H_0\), is that the increase in the substrate concentration does not cause a significant change in the mean velocity. The alternative hypothesis, denoted \(H_1\), is that the increase in concentration causes an increase in the mean velocity by more than 0.5 micromole per 30 minutes. Mathematically, this can be represented as: \(H_0: \mu_2 - \mu_1 \leq 0.5\) and \(H_1: \mu_2 - \mu_1 > 0.5\), where \(\mu_1\) and \(\mu_2\) are the mean velocities at 1.5 moles per liter and 2.0 moles per liter concentrations respectively.

02

Compute the pooled variance

The pooled variance is a weighted average of the variances of two samples and is calculated as follows: \(s_p^2 = \frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}\), where \(n_1\) and \(n_2\) are the sample sizes, and \(s_1\) and \(s_2\) are the standard deviations. Substitute the given values into the formula to get the pooled variance.

03

Compute the test statistic

The test statistic for the differences of means assuming equal variances is computed as follows: \(t = \frac{\bar{x}_2-\bar{x}_1-d_0}{s_p \cdot \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\), where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(d_0\) is the difference under the null hypothesis (0.5 in this problem), and \(s_p\) is the pooled standard deviation. Again, substitute the given values into the formula to get the test statistic.

04

Determination of the critical value and decision

To make a decision whether to reject the null hypothesis or not, we need to compare our test statistic with the critical value from the t-distribution table at the 0.01 level of significance. The degrees of freedom is calculated as \(df = n_1 + n_2 - 2\). If our computed t-value is greater than the critical value, we reject the null hypothesis, otherwise we fail to reject it. Make this comparison and finalize your decision regarding the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pooled Variance

In hypothesis testing, the concept of pooled variance is crucial when comparing two sample means, especially when these samples are assumed to have the same variance. Pooled variance is an estimate of the common variance between the two samples. It is a weighted average of the variances from both samples. The formula to find the pooled variance is:\[ s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} \]Here, \(n_1\) and \(n_2\) represent the sample sizes, while \(s_1\) and \(s_2\) stand for the sample standard deviations.
This formula allows us to account for the different sample sizes, giving more weight to the variance from the larger sample.

• Pooled variance gives a more accurate estimate of variance when sample sizes are unequal yet assume equal population variances.
• It helps in computing the standard error of the mean difference, which is vital for the next step in hypothesis testing.

t-Distribution

The t-distribution is fundamental in hypothesis testing when the sample size is small or when the population variance is unknown.
It is used instead of the normal distribution because it better accommodates the additional variability in the data when sample sizes are modest. When comparing two means with the t-distribution, the formula for the test statistic is:\[ t = \frac{\bar{x}_2-\bar{x}_1-d_0}{s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \]Here, \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(d_0\) is the hypothesized difference between the population means, and \(s_p\) is the pooled standard deviation.

• The t-distribution takes into account the degrees of freedom, which is calculated by \(df = n_1 + n_2 - 2\).
• It becomes very similar to the normal distribution as the sample size increases.
• It is pivotal for determining the critical value needed to make decisions about the null hypothesis.

Significance Level

The significance level, denoted by \(\alpha\), is a threshold used in hypothesis testing to decide whether an observed effect is statistically significant.In this problem, the significance level is set at 0.01, which means there is a 1% risk of rejecting the null hypothesis when it is actually true.

• A lower \(\alpha\) value indicates a stricter criterion for significance, reducing the risk of Type I errors (false positives).
• Common significance levels are 0.05, 0.01, and 0.10, with 0.01 being quite stringent.
• The chosen significance level affects the critical value from the t-distribution table against which the test statistic is compared.

In essence, the significance level helps in determining the robustness of the conclusions drawn from the hypothesis test.

Null and Alternative Hypothesis

In hypothesis testing, setting up null and alternative hypotheses is one of the first and critical steps.The null hypothesis \(H_0\) assumes that any observed difference is due to chance and not a systematic effect. For our specific example, \(H_0: \mu_2 - \mu_1 \leq 0.5\). This means the increase in substrate concentration does not cause a mean velocity increase greater than 0.5 micromoles per 30 minutes.The alternative hypothesis \(H_1\) suggests that there's an actual effect or difference. In our case, \(H_1: \mu_2 - \mu_1 > 0.5\), indicating the concentration indeed increases the velocity significantly.

• The null hypothesis is tested for rejection, being the default assumption.
• Rejecting \(H_0\) implies accepting \(H_1\), indicating sufficient evidence of a real effect.
• These hypotheses guide the direction of the test: right-tailed, left-tailed, or two-tailed.

Formulating the correct hypotheses is crucial as they lay the groundwork for all subsequent steps in hypothesis testing.