Multiply (5x-1)(6x^2+3x+7) SHOW WORK
2 answers:
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Answer:
Null hypothesis:
Alternative hypothesis:
with the excel code:"=T.INV(0.95,35)"
with the excel code:"=T.INV(0.90,35)"
If we compare the p value and the significance level given we see that
so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.
Step-by-step explanation:
Data given and notation
represent the mean time for the sample
represent the sample 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 mean is higher than 6 days, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(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".
Calculate the statistic
We can replace in formula (1) the info given like this:
Critical value and P-value
The first step is calculate the degrees of freedom, on this case:
In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates on the right. Using the significance level of 0.05 we got:
with the excel code:"=T.INV(0.95,35)"
And using the significance of 0.1 we got
with the excel code:"=T.INV(0.90,35)"
Since is a one side right tailed 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 reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.