Answer:
The period of the function is
. Then three times this period is
.
Step-by-step explanation:
Using Mathematica you can use the command Plot as follows:
```
Plot[Sin[x], {x,0,6Pi}]
```
The output is the graph shown below.
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Answer:
a) The hypothesis that the mean annual income is the same could not be rejected. There is no enough evidence to claim they are different.
b) The critical values for this two-sided test are t=±1.711.
Step-by-step explanation:
<em>The question is incomplete:</em>
<em>(a) Identify the claim and state H0 and Ha. </em>
<em> Which is the correct claim below?</em>
<em>(b) Find the critical value(s) and identify the rejection region(s). </em>
<em> Enter the critical value(s) below.</em>
We have an hypothesis test on the difference of two means.
The null and alternative hypothesis are:
The claim of the alternative hypothesis is that the mean annual income is different in County A and County B.
The null hypothesis is that the mean annual income is equal in both counties.
The significance level is α=0.10.
The sample from County A has a mean of S40,400, a s.d. of $8,700 and a sample size of 18 residents.
The sample from County B has a mean of S39,200, a s.d. of $6,000 and a sample size of 8 residents.
The standard error of the difference of means is:
The degrees of freedom are:
Then, the test statistic is:
For a statistic t=0.407, and df=24, the P-value is P=0.69. As the P-value is bigger than the significance level, the null hypothesis failed to be rejected.
If we would use the critial value approach, we would have to calculate the critical values for t, for df=24, two sided test and α=0.10.
The critical values, looking in a table, are t=1.711.