1 a) A random sample of 25 was drawn from a normal distribution with a standard deviation of 5. The sample mean is 80. Determine the 95% confidence interval estimate of the population mean. b) Repeat part (a) with a sample size of 100.c) Repeat part (a) with a sample size of 400.d) Describe what happens to the confidence interval estimate when the sample size increases.2 a) 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. Estimate the population mean with 99% confidence. b) Repeat part (a) changing the population standard deviation to 30.c) Repeat part (a) changing to the population standard deviation to 60.d) Describe what happens to the confidence interval estimate when the standard deviation is increased.

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.

i want to die

what the hell man, i just want the points lol