Answer To: 1 Math 107 – Modern Elementary Statistics Extra Credit/Remedial Work Instruction: Answer all...
Suraj answered on Aug 08 2021
1
Math 107 – Modern Elementary Statistics
Extra Credit/Remedial Work
Instruction: Answer all questions! Show your work properly!
PART 2
Name of Student: ____________________________________________________
1. (8 Points) Find the critical value Zα/2 corresponding to the given confidence level.
Confidence Level
α
Critical Value, Zα/2
92%
95%
94%
98%
Answers:
Confidence Level
α
Critical Value, Zα/2
92%
0.08
1.75
95%
0.05
1.96
94%
0.06
1.88
98%
0.02
2.326
Show your work here.
We can get the alpha,
a
value by subtracting the confidence level from 1. The calculations are given as follows:
92%0.92
10.92
0.08
Confidencelevel
a
==
=-
=
All the alpha values are calculated in same way.
To calculate the critical value, we need the normal distribution table or we can use MS-Excel formula.
For
0.08
a
=
,
0.08/2
NORM.S.INV(0.08/2)
z
=
1.75
=
In the same way we can calculate critical values by just changing the alpha value in the above formula.
2. (5 Points) The Medical Rehabilitation Education Foundation (MREF) reports that the average cost of rehabilitation for stroke victims is $24,672. To see if the average cost of rehabilitation is different at a particular hospital, a researcher selects a random sample of 35 stroke victims at the hospital and finds that the average cost of their rehabilitation is $25,226. The standard deviation of the population is $3251. At α = 0.01, can it be concluded that the average cost of stroke rehabilitation at a particular hospital is different from $24,672?
Answers:
Consider the given information,
The average cost of rehabilitation for stroke victims,
24672
m
=
The sample size, n = 35
The sample mean,
25226
x
=
The population standard deviation,
3251
s
=
The level of significance,
0.01
a
=
The hypotheses are given as follows:
0
1
:24672
:24672
H
H
m
m
=
¹
The test-statistic is given as follows:
/
x
z
n
m
s
-
=
(Because population standard deviation is given)
2522624672
3251/35
1.008
z
-
=
=
Since, the test is a two-tailed test. Thus, the p-value is calculated as follows:
(
)
21.008
20.1567
0.3134
pvaluePz
-=´>
=´
=
Since, the p-value is greater than the level of significance (0.3134 > 0.05). Thus, the null hypothesis is not rejected. Hence, we do no have enough evidence against the alternative hypothesis.
3. (8 Points) A popular social media platform claims that on an average each of its registered users logs into their profiles for about 30 times in a year. If the number of login data for a random sample of 144 users is collected and analyzed, it is found that the average login count in a year is 29. Assume that the population standard deviation is 3.1
Calculate the 95% confidence interval for population mean using the sample mean. Is the company’s claim is believable?
Answer(s):
The sample size, n = 144
The sample mean,
29
x
=
The population standard deviation,
3.1
s
=
The significance level,
0.05
a
=
The
/2
z
a
critical value from normal distribution table is 1.96.
The formula for the confidence interval is given as follows:
/2/2
,
xzxz
nn
aa
ss
æö
-+
ç÷
èø
Substitute the values in the above formula,
(
)
3.13.1
291.96,291.96
144144
29.49,29.51
æö
-´+´
ç÷
èø
Thus, 95% confidence interval is 29.49 and 29.51. The confidence interval does not contain the true population mean 30. Thus, the company claim is claim is correct.
4. (5 Points) Assume the time taken by racers to finish the running race (100 mts) follows a normal distribution. If you want to estimate the average time taken by racers to finish the 100 mts race with 90% confidence level and you want to express the interval within 2 secs, what is the minimum sample size that you need to consider? From the past studies the standard deviation for time taken by racers is 2.1 secs.
Answer(s):
The objective is to calculate the sample size.
The information given is as follows:
Confidence level = 90%
Significance level,
10.90
a
=-
0.10
=
0.10/2
1.645
z
=
The population standard deviation, s = 2.1
Margin of error = 2
The required sample size,
2
/2
z
n
a
s
e
´
æö
=
ç÷
èø
2
1.6452.1
2
3
´
æö
=
ç÷
èø
;
5. (5 Points) A research firm wants to estimate the average time (Industry average) of a client call. A random sample of 60 calls made by sales teams from different companies is analyzed. Average time for the sample is found to be 3.4 minutes. If the standard deviation of population is known to be 0.4 minutes, Estimate the 99% confidence interval for the average time of client call and interpret this interval.
Answer(s):
The sample size, n = 60
The sample mean,
3.4
x
=
The population standard deviation,
0.4
s
=
The significance level,
0.01
a
=
The
/2
z
a
critical value from normal distribution table is 2.576.
The formula for the confidence interval is given as follows:
/2/2
,
xzxz
nn
aa
ss
æö
-+
ç÷
èø
Substitute the values in the above formula,
(
)
0.40.4
3.42.576,3.42.576
6060
3.27,3.53
æö
-´+´
ç÷
èø
Thus,...