Question

1) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p
n = 110, x=55, 88% confidence
a. 0.426 < p < 0.574
b. 0.422 < p < 0.578
c. 0.425 < p < 0.575
d. 0.421 < p < 0.579

2) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p
A study involves 669 randomly selected deaths, with 31 of them caused by accidents. Construct a 98% confidence interval for the true percentage of all deaths that are caused by accidents.
a. 3.29% < p < 5.97%
b. 3.04% < p < 6.23%
c. 2.74% < p < 6.53%
d. 2.54% < p < 6.73%

3) Use the confidence level and sample data to find a confidence interval for estimating the population μ. Round your answer to the same number of decimal places as the sample mean.
A random sample of 130 full-grown lobsters had a mean weight of 21 ounces and a standard deviation of 3.0 ounces. Construct a 98% confidence interval for the population mean μ.
a. 20 oz < μ < 22 oz
b. 21 oz < μ < 23 oz
c. 19 oz < μ < 21 oz
d. 20 oz < μ < 23 oz

4) Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
n = 30, x = 83.1, s = 6.4, 90% confidence
a. 80.71 < μ < 85.49
b. 79.88 < μ < 86.32
c. 81.11 < μ < 85.09
d. 81.13 < μ < 85.07

5) Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
Thirty randomly selected students took the calculus final. If the sample mean was 83 and the standard deviation was 13.5, construct a 99% confidence interval for the mean score of all the students.
a. 76.93 < μ < 89.07
b. 76.21 < μ < 89.79
c. 78.81 < μ < 87.19
d. 76.23 < μ < 89.77

6) Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ. Assume that the population has a normal distribution.
A savings and loan association needs information concerning the checking account balances of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking account balance for local customers.
a. $455.65 < μ < $872.63
b. $492.52 < μ < $835.76
c. $493.71 < μ < $834.57
d. $453.59 < μ < $874.69

7) Use the given degree of confidence and sample data to construct a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
College students' annual earnings: 98% confidence interval; n = 9, x = $3959, s = $886
a. $539 < σ < $1734
b. $598 < σ < $1697
c. $697 < σ < $1156
d. $559 < σ < $1953

8) 41 random samples of monthly electric bill amounts are selected form a normally distributed population. The samples have a mean of $108 and a standard deviation of $5. Construct a 98% confidence interval for the population standard deviation.
a. $1.89 < σ < $2.75
b. $3.14 < σ < $9.02
c. $3.96 < σ < $6.72
d. $4.23 < σ < $6.14

9) Use the given degree of confidence and sample data to construct a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
Heights of mature White Pine trees: 99% confidence; n = 29, x = 65 ft, s = 8.4 ft
a. 8.27 ft < σ < 8.53 ft
b. 4.23 ft < σ < 10.47 ft
c. 6.22 ft < σ < 12.59 ft
d. 5.73 ft < σ < 10.07 ft

10) Use the given degree of confidence and sample data to construct a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.
The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.9 15.2 15.3 15.2
15.3 15.8 15.1 15.2
Find a 98% confidence interval for the population standard deviation σ.
a. 0.18 oz < σ < 0.62 oz
b. 0.19 oz < σ < 0.62 oz
c. 0.21 oz < σ < 0.80 oz
d. 0.19 oz < σ < 0.72 oz

Answer 1

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

Explanation:

The Confidence Interval, CI are given as follows

For the population proportion

`CI=\hat{p}\pm z* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}`

1)
`\hat p` = 55/110 = 0.5

z at 88% = ±1.56

a. CI = 0.426 < p < 0.574

2)
`\hat p` = 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

`\bar x` = 21

n = 130

s = 3.0

Confidence level = 98%

`CI=\bar{x}\pm t_(\alpha/2) (s)/(√(n))`

`t_(\alpha /2)` = ±2.356

CI = 20.380 < μ < 21.62

Rounding up gives;

a. CI = 20 < μ < 22

4) Similarly here we have;

`t_(\alpha /2)` = ±1.699 and

c. CI = 81.11 < μ < 85.09

5) Here

`\bar x` = 83

s = 13.5

n = 30

Confidence level = 99%

`t_(\alpha /2)` = ±2.76 and

b. CI = 76.21 < μ < 89.79

6) In the question, we have

`\bar x` = $664.14

s = $297.29

n = 14

Confidence level = 98%

`t_(\alpha /2)` = ±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;

`\sqrt{\frac{\left (n-1 \right )s^(2)}{\chi _(1-\alpha /2)^{}}}< \sigma < \sqrt{\frac{\left (n-1 \right )s^(2)}{\chi _(\alpha /2)^{}}}`

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.