Question

2. It is well known that astronauts increase their height in space missions because of the lack of gravity. A question is whether or not we increase height here on Earth when we are put
into a situation where the effect of gravity is minimized. In particular, do people grow taller when confined to a bed? A study was done in which the heights of six men were taken before
and after they were confined to bed for three full days.
a. The before-after differences in height measurements (in millimeters) for the six men were
12.6 14.4 14.7 14.5 15.2 13.5.
Assuming that the men in this study are representative of the population of all men, what is an estimate of the population mean increase in height after three full days in bed?
b. Calculate the margin of error associated with your estimate of the population mean from part (a). Round your answer to three decimal places.
c. Based on your sample mean and the margin of error from parts (a) and (b), what are plausible values for the population mean height increase for all men who stay in bed for three full days?

Answer 1

Answer:

a) $\bar d = \frac{12.6 +14.4 +14.7 +14.5 +15.2 +13.5}{6}=14.15$

b) $ME=2.57 \frac{0.940}{\sqrt{6}}=0.986$

c) $14.15 - 2.57 \frac{0.940}{\sqrt{5}}=13.164$

$14.15 + 2.57 \frac{0.940}{\sqrt{5}}=15.136$

The 95% confidence interval is given by (13.164.15.136)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Part a

12.6 14.4 14.7 14.5 15.2 13.5.

Assuming that the men in this study are representative of the population of all men, what is an estimate of the population mean increase in height after three full days in bed?

For this case the best estimate for the mean is the average given by this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

For our case we are taking differences so would be the mean of differences and we got:

$\bar d = \frac{12.6 +14.4 +14.7 +14.5 +15.2 +13.5}{6}=14.15$

Part b

Assuming 95 % of confidence level. In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. The degrees of freedom are given by:  

$df=n-1=6-1=5$  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,5)" for $t_{\alpha/2}=-2.57$

"=T.INV(1-0.025,5)" for $t_{1-\alpha/2}=2.57$  

The critical value $tc=\pm 2.57$

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$ME=t_c \frac{s_d}{\sqrt{n}}$

First we calculate the sample deviation for the differences with this formula:

$s_d = \sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}}=0.940$

$ME=2.57 \frac{0.940}{\sqrt{6}}=0.986$

Part c

The interval for the mean is given by this formula:

$\bar d \pm t_{c} \frac{s_d}{\sqrt{n}}$

And calculating the limits we got:

$14.15 - 2.57 \frac{0.940}{\sqrt{5}}=13.164$

$14.15 + 2.57 \frac{0.940}{\sqrt{5}}=15.136$

The 95% confidence interval is given by (13.164.15.136)