Mathematics, 18.12.2019 00:31 mandyj5477

Consider a chemical company that wishes to determine whether a new catalyst, catalyst xa-100, changes the mean hourly yield of its chemical process from the historical process mean of 750 pounds per hour. when five trial runs are made using the new catalyst, the following yields (in pounds per hour) are recorded: 801, 814, 784, 836, and 820. chemyield

t-statistic
6.942585
p-value
0.002261

Answers: 3

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Answer from: alyssabalinnt

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

$p_v =2*P(t_{4}6.943)=0.00226$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is significantly different from 750 pounds per hour.

Step-by-step explanation:

Data given and notation

Data: 801, 814, 784, 836,820

We can calculate the sample mean and sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=811$ represent the sample mean

s=19.647 represent the standard deviation for the sample

n=5 sample size

$\mu_o =750$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is different from 750 pounds per hour, the system of hypothesis would be:

Null hypothesis: $\mu = 750$

Alternative hypothesis: $\mu \neq 750$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{811-750}{\frac{19.647}{\sqrt{5}}}=6.943$

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

What do you conclude?

Compute the p-value

Since is a two tailed test the p value would be:

$p_v =2*P(t_{4}6.943)=0.00226$

Answer from: Quest

answer is : muchas/tipicas

have a nice day!

~recklesschicken

Answer from: Quest

step-by-step explanation:

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Consider a chemical company that wishes to determine whether a new catalyst, catalyst xa-100, change...

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