confidence intervals questions

Question

the two sample are

Female height : 101,102,102,104,107,107,110,110,112,115,122,123,123,123,126,127,129,133,133,138,140,144,144,146,148,149,150,153,153,159,159,162,164,164,165,169,173,174,174,176

Make height: 107,108,110,111,111,115,121,123,125,129,132,134,135,136,136,139,140,141,142,149,153,154,156,161,163,163,164,165,166,167,169,169,171,172,178,179,181,182,182,183

1. calculate for each sample ( males and females separately ) :
a.) The mean and standard deviation of heights. There should be two means and two standard deviations.
b.) Two 95% confidence intervals for the mean heights ( males and females)

2. FULLY EXPLAIN the meaning of the 95% confidence intervals for the mean height for males and females.

3. For height, ( males and females), calculate the height above which 20% of the height are recorded.

4. Use your knowledge of "confidence intervals for the difference of two mean" to investigate the difference between the mean heights of the two populations "males" and "females". Both confidence interval levels 95% and 99% should be used.

5. Is there any significant between the mean heights of the males and the females? Justify your answer for both confidence level 95% and 99%. Use your confidence intervals in question 4 as evidence and any other data you feel is relevant.

thanks people

Answer 1

Thank you MS Excel... text to columns....
did not have to type a single number...
copy and paste into Excel... text to columns...

PART I:

Ladies first:
the mean is 137.825
the standard deviation is 23.672

Now for the guys:
the mean is 148.05
the standard deviation is 23.9572

The confidence interval formula is

mean +or- 1.96*standard deviation/sqrt(n)


There are 40 statistics and 1.96 comes from
the 95% condfidence interval assuming normal distribution.

For the ladies (130.489, 145.161)

For the guys: (140.6256,155.4744)


PART 2:

The 95% confidence interval means that
we are 95% certain that the REAL POPULATION
mean lies inside that interval.


Part 3:

Let z be the height at which 20% = 0.2 of the statistics are greater than
or equal to z.

That is, Prob(Z > z) = 0.2

This probability is the same as Prob( Z <= z) = 0.8
That means 80% of the statistics are Less than this missing number.

On the normal curve, per the normal table, the z value is 0.841

The formula for converting to z-score is:
z = (x - mean)/ standard deviation

Solving for x:

z * standard deviation + mean = x

For the ladies: x = 157.733
For the guys: x=168.2


Part 4:

A requirement for this type of confidence interval is
that the population variances are the same.

Based on the sample variances, it appears that this is the case.

The formula is :

Lower limit: (Mean 1 - Mean 2) - z* MSE
Upper limit: (Mean 1 - Mean 2) + z * MSE

where MSE is the mean-square-error = sqrt( 2* standard deviation)^2/n)

For 95% confidence, z = 1.95
For 99% confidence, z = 2.58

Will use (23.67178+23.95717)/2 = 23.814475 for the population standard deviation.

The mean square error is sqrt( 2 * 23.814475^2/40)=5.3251

The difference in the means is 10.225

95% ---> (-0.212196 , 20.662196)

99% ----> (-3.513758 , 23.963758)

SO we are 99% certain that the actual difference between the two
population means lies is no larger than 23.963758 and no smaller
than -3.513758