1) a. CI = 0.426 < p < 0.574

2) c. CI = 2.74% < p < 6.524%

3) a. CI = 20 < μ < 22

4) c. CI = 81.11 < μ < 85.09

5) b. CI = 76.21 < μ < 89.79

6) d. $453.59 < μ < $874.69

7) d. CI = $559 < σ < $1953.3

8) c. CI = $3.96 < σ < $6.72

9) c. 6.22 ft < σ < 12.59 ft

10) d. 0.19 oz < σ < 0.72 oz

Step-by-step explanation:

The Confidence Interval, CI are given as follows

For the population proportion

1) = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2) = 31/669 = 0.5

z at 98% = ±2.326

c. CI = 2.74% < p < 6.524%

3) Here we have unknown population standard deviation, so we find the t interval

= 21

n = 130

s = 3.0

Confidence level = 98%

= ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

= ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

= 83

s = 13.5

n = 30

Confidence level = 99%

= ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

= $664.14

s = $297.29

n = 14

Confidence level = 98%

= ±2.65 and

CI = $453.56 < μ < $874.72

d. $453.59 < μ < $874.69

7) The confidence interval for a population standard deviation is given by the following relation;

Here

n = 9

x = $3959

s = $886

Therefore we have

d. CI = $559 < σ < $1953.3

8) Here we have;

n = 41

x = $108

s = $5

Therefore;

c. CI = $3.96 < σ < $6.72

9) Here we have;

n = 29

x = 65 ft

s = 8.5 ft

Therefore we have

CI = 6.3 ft < σ < 12.73 ft

c. 6.22 ft < σ < 12.59 ft

10) Here we have

n = 8

s = 0.301188123

d. 0.19 oz < σ < 0.72 oz.