Which type of question leads a person to a particular answer? A.honest, B. fair, C. friendly, D. biased

(1 point) A recent report for a regional airline reported that the mean number of hours of flying time for its pilots is 56 hour

Ray Of Light [21]

Answer:

(53.812 ; 58.188) ; 156

Step-by-step explanation:

Given that :

Sample size (n) = 51

Mean (m) = 56

Standard deviation (σ) = 9.5

α = 90%

Using the relation :

Confidence interval = mean ± Error

Error = Zcritical * (standard deviation / sqrt (n))

Zcritical at 90% = 1.645

Error = 1.645 * (9.5 / sqrt(51))

Error = 1.645 * 1.3302660

Error = 2.1882877

Hence,

Confidence interval :

Lower boundary = 56 - 2.1882877 = 53.8117123

Upper boundary = 56 + 2.1882877 = 58.1882877

Confidence interval = (53.812 ; 58.188)

2.)

Margin of Error (ME) = 1.25

α = 90%

Sample size = ((Zcritical * σ) / ME)^2

Zcritical at 90% = 1.645

Sample size = ((1.645 * 9.5) / 1.25)^2

Sample size = (15.6275 / 1.25)^2

Sample size = 12.502^2 = 156.3000

Sample size = 156

3 0

2 years ago

Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b

kramer

Answer:

(a)

$\bar x = 0.0075$

$s = 0.0049$

(b)

B. The sample is too small to make judgments about skewness or symmetry.

Step-by-step explanation:

Given:

$n = 8$

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 \ \end{array}$

Solving (a):

First, calculate the difference between the recorded TBBMC for both operators:

$\begin{array}{ccccccccc}{} & {S} & {u} & {b} &{j} & {e} & {c} & {t} & {s} &{Operator} & {1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {1} & 1.326 & 1.337 & 1.079 & 1.229 & 0.936 & 1.009 & 1.179 & 1.289 & 2 & 1.323 & 1.322 & 1.073 & 1.233 & 0.934 & 1.019 & 1.184 & 1.304 &{|1 - 2|} &0.003 & 0.015 & 0.006 & 0.004 & 0.002 & 0.010 & 0.005 & 0.015 \ \end{array}$

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.

The mean is then calculated as:

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{0.003 + 0.015 + 0.006 + 0.004 + 0.002 + 0.010 + 0.005 + 0.015}{8}$

$\bar x = \frac{0.06}{8}$

$\bar x = 0.0075$

Next, calculate the standard deviation (s).

This is calculated using:

$s = \sqrt{\frac{\sum (x - \bar x)^2}{n}}$

So, we have

$s = \sqrt\frac{(0.003 - 0.0075)^2 + (0.015 - 0.0075)^2+ (0.006- 0.0075)^2+ ....... + (0.005- 0.0075)^2+ (0.015- 0.0075)^2 }{8}$ $s = \sqrt\frac{0.00019}{8}$