Find the circumference of a circle with a diameter 1 1/2 inches
1 answer:
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Answer:
Hence the expression
Step-by-step explanation:
Explanation
- The given expression is (a + b)(2a+3 b).
- We have to multiply the given expression.
- Multiply the (a + b) by 3b, multiply the (a + b) by 2a then add like terms.
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Answer:
b. H0: μ < 530. Reject H0 if tcalc > -1.753
Step-by-step explanation:
1) Data given and notation
represent the sample mean
represent the sample standard deviation
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 less than 530 (left tailed tes), the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and 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
Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:
Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be .
And we can use the following excel code to find it: "=T.INV(0.05,15)"
So then the correct rejection zone for H0 would be: Reject H0 if
P-value
Since is a one side left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level for example we see that
so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.