The radius of the circle is 4 inch and has an arc length of 5.8 inches . We know that if the central angle is 360° then the arc length is . Similarly when angle is theta ,
Now substitute respective values , from the given data ,
This is our required answer!
Answer:
(A) A 95% confidence for the population mean is [$332.16, $447.84] .
(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.
(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.
(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .
Step-by-step explanation:
We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~
where, = sample mean = $390
s = sample standard deviation = $120
n = sample of students = 19
= population mean
<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>
<u>So, 95% confidence interval for the population mean, </u><u> is ;
</u>
P(-2.101 < < 2.101) = 0.95 {As the critical value of t at 18 degrees of
freedom are -2.101 & 2.101 with P = 2.5%}
P(-2.101 < < 2.101) = 0.95
P( <
<
) = 0.95
P( <
<
) = 0.95
<u>
95% confidence interval for</u> = [
,
]
= [ ,
]
= [$332.16, $447.84]
(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .
(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is would increase because of an increase in the z value.
(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is would decrease because as denominator increases; the whole fraction decreases.
(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion students who purchase their used textbooks =
= 0.42
n = sample of students = 500
p = population proportion
<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions</em>
<u>So, 99% confidence interval for the population proportion, p is ; </u>
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%
level of significance are -2.58 & 2.58}
P(-2.58 < < 2.58) = 0.99
P( <
<
) = 0.99
P( < p <
) = 0.99
<u>
99% confidence interval for</u> p = [ ,
]
= [ ,
]
= [0.363, 0.477]
Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .