Answered

Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

. A study was conducted 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 s substrate concentration of 2.0 moles per liter, 12 runs were made, yielding 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 of the reaction of 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.

Sagot :

contexto1024

At the significance level, 0.01

p-Value=0.973 > 0.01 which implies fail to reject null hypothesis and no evidence to support the claim.

So, there is no reason to believe that this increase in substrate concentration causes an increase in the mean velocity of the reaction of more than 0.5 micromole per 30 minutes.

We have given that,

A study conducted to see if increasing the substrate concentration has an appreciable effect on the velocity of a chemical reaction.

x₁-bar = 8.8 micromoles

Standaard deviations, s₁= 1.2

x₂-bar = 7.5 ; s₂ = 1.5

n₁ = 12 ; n₂ = 15

significance level = 0.01

Consider the Null hypothesis and alternative hypothesis othesis

H₀ : μ = 0.5

Hₐ : μ > 0.5

Consider,

Pooled Standard Deviation = √(((n₁ - 1)s₁² + (n₂- 1)s₂²)/(n₁ + n₂-2))

plugging all known values

= sqrt(((12-1)(1.2²) + (15 -1)(1.5²))/(12+15-2))

Sp = 1.3761

Now , test statistic :

t =((x₁-bar - x₂-bar) - 0.5) / (Sp×√(1/n₁+ 1/n₂))

substitute all known values we get,

= ((8.8 - 7.5) - 0.5) / (1.3761× sqrt(1/12 + 1/15))

=> t = 1.5

Degree of freedom = n₁+ n₂ - 2

= 12 + 15 - 2

= 25

Now the p-value corresponding to t = 1.5 & df = 25 is 0.073

Here p-value > 0.01 (significance level) thus, we fail to reject null hypothesis.

So there is no enough evidence to support the claim.

To learn more about Hypothesis testing, refer:

https://brainly.com/question/4232174

#SPJ4

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.