What is the probability of obtaining prime numbers when a fair dice is thrown?
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Answer:
A 90% confidence interval for <em>p</em> will be <u>narrower </u>than the 99% confidence interval.
Step-by-step explanation:
The formula to compute the (1 - <em>α</em>) % confidence interval for a population proportion is:
Here is the sample proportion.
The margin of error of the confidence interval is:
The MOE is dependent on:
- Confidence level
- Standard deviation
- Sample size
The MOE is directly related to the confidence level and standard deviation.
So if any of the two increases then the MOE also increases, thus widening the confidence interval.
And the MOE is inversely related to the sample size.
So if the sample increases the MOE decreases and vice versa.
It is provided that the sample size and the sample proportion are not altered.
The critical value of <em>z</em> for 90% confidence level is:
And the critical value of <em>z</em> for 99% confidence level is:
So as the confidence level increases the critical value increases.
Thus, a 90% confidence interval for <em>p</em> will be narrower than the 99% confidence interval.
We have that every quarter, he will get an 0.8% interest rate. We also have that there will be 14*4=56 periods of repayment. We have that by the formula, if there are n periods and x% is the interest rate, the capital is multiplied by (100+x%)^n.
Suppose Micah invests x money. Then, we have that after 14 years, the capital available will be C=
. We have that C=18500$ and hence that 18500=x*1.562. Thus, solving for x we have that x=11841$.
Micah will need to invest only 11841$ in order to pay for his boat; this shows the importance of saving since he earns a lot of money, around 7k $.
Suppose Micah invests x money. Then, we have that after 14 years, the capital available will be C=
Micah will need to invest only 11841$ in order to pay for his boat; this shows the importance of saving since he earns a lot of money, around 7k $.