In: Statistics and Probability
Problem: Construct and interpret a 90%, 95%, and 99% confidence interval for the mean heights of either adult females or the average height of adult males living in America. Do not mix genders in your sample as this will skew your results. Gather a random sample of size 30 of heights from your friends, family, church members, strangers, etc. by asking each individual in your sample his or her height. From your raw data convert individual heights to inches. Record your raw data and your conversions in the table on page 2 of this document. Construct and interpret the confidence interval based on the raw data from your random sample. In a word processed document, answer the reflections questions below. Use the equation editor to show your calculations for the percent difference indicated in 6) below.
Reflections: 1) Summarize the characteristics of your sample – how many was in it, who was in it, from where did you get your sample, what would you estimate to be the average age of your sample, etc.?
2) What is x for your sample?
3) What is s for your sample?
3) State and interpret the 90% confidence interval for your sample.
4) State and interpret the 95% confidence interval for your sample.
5) State and interpret the 99% confidence interval for your sample.
6) Research from a credible source the average height in the population as a whole for the group you sampled. Make sure to credit your source. Calculate a percent difference between the average of your sample and the average in the population as a whole. What was the percent difference of the average height in your sample and the population as a whole? Comment on your percent difference.
Table of Raw Data of womens heights
Sample Number |
Height in Feet and Inches |
Height Converted to Inches |
1 |
5 feet |
60 |
2 |
5 feet 3 inches |
63 |
3 |
5 feet 5 inches |
65 |
4 |
5 feet 5 inches |
65 |
5 |
5 feet 9 inches |
69 |
6 |
5 feet 11 inches |
71 |
7 |
5 feet 1 inch |
61 |
8 |
5 feet 2 inches |
62 |
9 |
5 feet 3 inches |
63 |
10 |
5 feet 6 inches |
66 |
11 |
6 feet |
72 |
12 |
5 feet 11 inches |
71 |
13 |
5 feet 4 inches |
64 |
14 |
5 feet 8 inches |
68 |
15 |
5 feet 8 inches |
68 |
16 |
5 feet 4 inches |
64 |
17 |
5 feet 7 inches |
67 |
18 |
5 feet 5 inches |
65 |
19 |
5 feet 5 inches |
65 |
20 |
5 feet 2 inches |
62 |
21 |
5 feet 5 inches |
65 |
22 |
5 feet 9 inches |
69 |
23 |
5 feet 2 inches |
62 |
24 |
5 feet 3 inches |
63 |
25 |
5 feet 1 inches |
61 |
26 |
5 feet 4 inches |
64 |
27 |
5 feet 5 inches |
65 |
28 |
5 feet 5 inches |
65 |
29 |
5 feet 3 inches |
63 |
30 |
5 feet 6 inches66 |
66 |
Solution1:
in excel
install anlaysis tool pack then go to data>data analysis>descriptive statistics.
select the data range
you will get
Height Converted to Inches | |
Mean | 65.13333 |
Standard Error | 0.566768 |
Median | 65 |
Mode | 65 |
Standard Deviation | 3.104317 |
Sample Variance | 9.636782 |
Kurtosis | -0.21166 |
Skewness | 0.600386 |
Range | 12 |
Minimum | 60 |
Maximum | 72 |
Sum | 1954 |
Count | 30 |
mean=65.13333
median=65
mode=65
mean=median =mode
shape :symmetrical
follows normal distribution.
No outliers found
Solution2:
x of the sample is sample mean=xbar=65.1333
Solution3:
s of the sample is standard deviation=s=
3.104317 |
Solution4:
alpha=1-0.90=0.1
90% confidence interval for your sample.
xbar-MOE,xbar+MOE
MOE=margin of error=tcrit *s/sqrt(n)
MOE can be found in excel as
syntax is CONFIDENCE.T(alpha.samplesd,samplesize)
=CONFIDENCE.T(0.1;3.104317;30)
=0.963011079
90% lower limit=65.1333-0.963011079=64.17029
90% upper limit=65.1333+0.963011079=66.09631
we are 90% confident that the true population mean height of womens height lies in between
64.17029 and 66.09631 inches