In a chemical compound there are three parts zinc for every 16 parts copper by mass. A piece of the compound contains 320 g of c
1 answer:
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Answer:
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the population standard deviation for the sample
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
(1)
z-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".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
Since is a two sided test the p value would be:
Conclusion
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance
Answer:
Step-by-step explanation:
Hello!
Given the variables
X: Curfew of middle school students (Categorized Yes/No)
Y: Average grade of middle school students. (Categorized: A, B, C, and D)
The objective is to test if there is an association between both variables, you have to conduct a Chi-Square test of independence.
In the null hypothesis, you state that both variables are independent vs. the alternative hypothesis that the variables are dependent.
The hypotheses for this test are:
H₀: Pij= Pi. * P.j i= (1)Yes, (2)No; j= (1)A, (2)B, (3)C, (4)D
H₁: Te variables are not independent.
α: 0.05
Where
Oij is the observed frequency for the i-row and j-column
Eij is the expected frequency for the i-row and j-column
r= number of categories in the rows
c= number of categories in the columns
p-value: 0.687
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. So using a 5% significance level, you can conclude that having a curfew and the average grades of middle schoolers are two independent variables.
