Answer:
1a)Confidence Interval: (77.936,82.064)
b)Confidence Interval: (79.020, 80.980)
c)Confidence Interval: (79.510, 80.490)
d) As the sample size increases, the confidence interval for the lower class (bound) increases while that of the upper class(bound) decreases.
2a)Confidence Interval: (491.609
, 502.797)
b)Confidence Interval: (483.218, 516.782)
c)Confidence Interval: (466.436,533.564)
d)As the standard deviation increases, the confidence interval for the lower class (bound) decreases while that of the upper class(bound) increases.
Step-by-step explanation:
1 a) A random sample of 25 was drawn from a normal distribution with a standard deviation of 5. The sample mean is 80.
The formula for confidence Interval is:
C.I = Mean ± z score × Standard deviation/√n
Note to apply the formula above n greater than or equal to 30
a) Determine the 95% confidence interval estimate of the population mean.
For a, the confidence interval formula above cannot be applied because n is less than 30, so we use the t score confidence interval value
C.I = Mean ± t score × Standard deviation/√n
Mean = 80
Standard deviation = 5
n = 25
Degree of freedom = 25 - 1 = 24
Hence, 95% confidence interval degree of freedom t score = 2.064
Hence, Confidence Interval =
80 ± 2.064 × 5/√25
80 ± 2.064 × 1
80 ± 2.064
Confidence Interval =
80 - 2.064
= 77.936
80 + 2.064
= 82.064
Confidence Interval: (77.936, 82.064)
b) Repeat part (a) with a sample size of 100.
C.I = Mean ± z score × Standard deviation/√n
Mean = 80
Standard deviation = 5
n = 100
z score for 95% confidence interval = 1.96
C.I = 80 ± 1.96 × 5/√100
C.I = 80 ± 1.96 × 5/10
C.I = 80 ± 1.96 × 0.5
C.I = 80 ± 0.980
Confidence Interval =
80 - 0.980
= 79.020
80 + 0.980
= 80.980
Confidence Interval: (79.020, 80.980)
c) Repeat part (a) with a sample size of 400.
C.I = Mean ± z score × Standard deviation/√n
Mean = 80
Standard deviation = 5
n = 400
z score for 95% confidence interval = 1.96
C.I = 80 ± 1.96 × 5/√400
C.I = 80 ± 1.96 × 5/20
C.I = 80 ± 1.96 × 0.25
C.I = 80 ± 0.490
Confidence Interval =
80 - 0.490
= 79.510
80 + 0.490
= 80.490
Confidence Interval: (79.510, 80.490)
d) As the sample size increases, the confidence interval for the lower class (bound) increases while that of the upper class(bound) decreases.
2a) The mean of a sample of 25 was calculated as ¯x=500. The sample was randomly drawn from a population with a standard deviation of 15.
For a,b and c the confidence interval formula above cannot be applied because n is less than 30, so we use the t score confidence interval value
C.I = Mean ± t score × Standard deviation/√n
a) Mean = 500
Standard deviation = 15
n = 25
Degree of freedom = 25 - 1 = 24
Hence, 99% confidence interval degree of freedom t score = 2.064
Hence, Confidence Interval =
500 ± 2.797 × 15/√25
500 ± 2.797 × 15/5
500 ± 2.797 × 3
Confidence Interval =
500 - 8.391
= 491.609
500 + 8.391
= 508.391
Confidence Interval: (491.609, 508.391)
b) Repeat part (a) changing the population standard deviation to 30
Hence, Confidence Interval =
500 ± 2.797 × 30/√25
500 ± 2.797 × 30/5
500 ± 2.797 × 6
500 ± 16.782
Confidence Interval =
500 - 16.782
= 483.218
500 + 16.782
= 516.782
Confidence Interval: (483.218, 516.782)
c) Repeat part(a) changing to the population standard deviation to 60.
Hence, Confidence Interval =
500 ± 2.797 × 60/√25
500 ± 2.797 × 60/5
500 ± 2.797 × 12
500 ± 33.564
Confidence Interval =
500 - 33.564
= 466.436
500 + 33.564
= 533.564
Confidence Interval: (466.436,533.564)
d)As the standard deviation increases, the confidence interval for the lower class (bound) decreases while that of the upper class(bound) increases.
