How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer? What is the role of gravity in maintaining the atmospheric layer around the earth ?
Please put your answers with clear explanation.
Thank you ! ​

) A B (ii) A E (iii) C D

(iv) D F (v) F A (vi) D E

(vii) F B

\bar x = 260.1615

x

ˉ

=260.1615

\sigma = 70.69σ=70.69

The confidence interval of standard deviation is: 53.7653.76 to 103.25103.25

Step-by-step explanation:

Given

n =20n=20

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

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

x

ˉ

=

n

∑x

So, we have:

\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 ++250.61}{20}

x

ˉ

=

20

242.87+212.00+260.93+284.08+194.19+139.16+260.76+436.72+355.36++250.61

\bar x = \frac{5203.23}{20}

x

ˉ

=

20

5203.23

\bar x = 260.1615

x

ˉ

=260.1615

\bar x = 260.16

x

ˉ

=260.16

Solving (b): The standard deviation

This is calculated as:

\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}σ=

n−1

∑(x−

x

ˉ

)

2

\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2++(250.61- 260.1615)^2}{20 - 1}}σ=

20−1

(242.87−260.1615)

2

+(212.00−260.1615)

2

+(260.93−260.1615)

2

+(284.08−260.1615)

2

++(250.61−260.1615)

2

\sigma = \sqrt{\frac{94938.80}{19}}σ=

19

94938.80

\sigma = \sqrt{4996.78}σ=

4996.78

\sigma = 70.69σ=70.69 --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

c =0.95c=0.95

So:

\alpha = 1 -cα=1−c

\alpha = 1 -0.95α=1−0.95

\alpha = 0.05α=0.05

Calculate the degree of freedom (df)

df = n -1df=n−1

df = 20 -1df=20−1

df = 19df=19

Determine the critical value at row df = 19df=19 and columns \frac{\alpha}{2}

2

α

and 1 -\frac{\alpha}{2}1−

2

α

So, we have:

X^2_{0.025} = 32.852X

0.025

2

=32.852 at \frac{\alpha}{2}

2

α

X^2_{0.975} = 8.907X

0.975

2

=8.907 --- at 1 -\frac{\alpha}{2}1−

2

α

So, the confidence interval of the standard deviation is:

\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} } to \sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }

70.69 * \sqrt{\frac{20 - 1}{32.852} to 70.69 * \sqrt{\frac{20 - 1}{8.907}

70.69 * \sqrt{\frac{19}{32.852} to 70.69 * \sqrt{\frac{19}{8.907}

53.7653.76 to 103.25103.25

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.