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Answer:
99% confidence interval is wider as compared to the 80% confidence interval.
Step-by-step explanation:
We are given that a magazine provided results from a poll of 500 adults who were asked to identify their favorite pie.
Among the 500 respondents, 14% chose chocolate pie, and the margin of error was given as plus or minus ±3 percentage points.
The pivotal quantity for the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of adults who chose chocolate pie = 14%
n = sample of adults = 500
p = true proportion
Now, the 99% confidence interval for p =
Here, = 1% so
= 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.
Also, Margin of error = = 0.03 for 99% interval.
<u>So, 99% confidence interval for p</u> =
= [0.14 - 0.03 , 0.14 + 0.03]
= [0.11 , 0.17]
Similarly, <u>80% confidence interval for p</u> =
Here, = 20% so
= 10%. So, the critical value of z at 10% significance level is 1.2816.
Also, Margin of error = = 0.02 for 80% interval.
So, <u>80% confidence interval for p</u> = [0.14 - 0.02 , 0.14 + 0.02]
= [0.12 , 0.16]
Now, as we can clearly see that 99% confidence interval is wider as compared to 80% confidence interval. This is because more the confidence level wider is the confidence interval and we are more confident about true population parameter.
- The first four terms are : 15000, 14850, 14700, 14550
- a = 15000 ; d = -150
- The explicit formula is :
The first term of the sequence, a = 15,000
Each subsequent winner gets $150 less
Hence, the common difference, d = - $150
The first four terms :
First term, a1 = $15000
Second term, a2 = a - d = $15000 - $150 = $14850
Third term, a3 = a2 - d = $14850 - $150 = $14700
Fourth term, a4 = a3 - d = $14700 - $150 = $14550
a, for the sequence is the first term = $15000
d, for the sequence is the common $150 (difference between successive wins).
Explicit formula for the general term :
For n = 1
Hence, the explicit formula for the nth term is :
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