For this case we must factor of the following expression:
By law of signs of multiplication it is fulfilled that:
So we have to:
Thus, by factoring we have:
Answer:
What is the equivalent expression for( 7 x 6+7 x y)
1 answer:
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Answer:
The rejection zone for this case would be:
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 true mean is significantly different from 75 at 5% of significance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the population standard deviation for the sample
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 different from 75, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case 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:
Cutoff for the rejection regon
Since our significance level is and we are conducting a bilateral test we need to find a quantile in the standard normal distribution that accumulates 0.025 of the area on each tail.
And for this case those values are
So the rejection zone for this case would be:
Our calculated value is not on the rejection zone. So we fail to reject the null hypothesis.
P-value
Since is a two sided test the p value would be given by:
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 true mean is significantly different from 75 at 5% of significance.