Question

hours	age	sex
7	17	f
8	40	f
8	60	m
5	35	f
6	22	m
4	7	m
7	16	f
6	10	m
4	19	f
5	9	m
6	14	m

Please make a confidence interval with the data given and give the mean, standard deviation, critical value, and the margin of error.

Second, for hours and age please give the Min, Q1, Med, Q3, Max, Mean and standard deviation (separate the hours and age so they have their own data set for each)

Please help!!

Answer 1

Solution:

For Hours:

By using R-software:

> lHours=c(7,8,8,5,6,4,7,6,4,5,6);Hours
[1] 7 8 8 5 6 4 7 6 4 5 6
> n1=length(Hours);n1
[1] 11
> summary(Hours)
Min. 1st Qu. Median Mean 3rd Qu. Max.
4 5 6 6 7 8
> mH=mean(Hours);mH #mean of Hours
[1] 6
> v=var(Hours);v # Sample variance
[1] 2
> s1=sqrt(v);s1 # Sample standard deviation
[1] 1.414214
> SDhr=sqrt(((n1-1)/n1)*v);SDhr # population standard deviation
[1] 1.3484

R-code:

Hours=c(7,8,8,5,6,4,7,6,4,5,6);Hours
n1=length(Hours);n1
summary(Hours)
mH=mean(Hours);mH #mean of Hours
v=var(Hours);v # Sample variance
s1=sqrt(v);s1 # Sample standard deviation
SDhr=sqrt(((n1-1)/n1)*v);SDhr # population standard deviation

From this R-Output of data set Hours we get,  

Min=4, Q1=5, Med=Q2=6, Q3=7, Max=8, Mean=6, Standard deviation=1.3484

Also,  

= 6, sl = 1.4142

using t- test we find 95 % C.I. as foloows,

$(\bar{x}- t_{(n-1),\alpha/2}*\frac{s_1}{\sqrt{n}},\bar{x}+ t_{(n-1),\alpha/2}*\frac{s_1}{\sqrt{n}})$

$(\bar{x}-M.E.,\bar{x}+ M.E.)$

Where,

Critical value:   $t_{(n-1),\alpha/2}=t_{(11-1),0.05/2}=t_{10,0.025}=2.228$ (From t-table)

By using MS-Excel Function "=TINV(0.05,10)")

$(6- 2.228*\frac{1.4142}{\sqrt{11}},6+2.228*\frac{1.4142}{\sqrt{11}})$

Margin of error=0.95

The required C.I. for Hour is

For Age:

> Age=c(17,40,60,35,22,7,16,10,19,9,14);Age
[1] 17 40 60 35 22 7 16 10 19 9 14
> n2=length(Age);n2
[1] 11
> summary(Age)
Min. 1st Qu. Median Mean 3rd Qu. Max.
7.00 12.00 17.00 22.64 28.50 60.00
> mA=mean(Age);mA #mean of Age
[1] 22.63636
> v1=var(Age);v1 # Sample variance
[1] 260.4545
> s2=sqrt(v1);s2 # Sample standard deviation
[1] 16.1386
> SDA=sqrt(((n2-1)/n2)*v);SDA # population standard deviation of Age
[1] 1.3484

R-Code:

Age=c(17,40,60,35,22,7,16,10,19,9,14);Age
n2=length(Age);n2
summary(Age)
mA=mean(Age);mA #mean of Age
v1=var(Age);v1 # Sample variance
s2=sqrt(v1);s2 # Sample standard deviation
SDA=sqrt(((n2-1)/n2)*v1);SDA # population standard deviation of Age

From this R-Output of data set Age we get,  

Min=7, Q1=12, Med=Q2=17, Q3=28.50, Max=60, Mean=22.64, Standard deviation=15.3876

Also,  $\bar{x}=6,s_2=16.1386$

using t- test we find 95 % C.I. as foloows,

$(\bar{x}- t_{(n-1),\alpha/2}*\frac{s_2}{\sqrt{n}},\bar{x}+ t_{(n-1),\alpha/2}*\frac{s_2}{\sqrt{n}})$

$(\bar{x}-M.E.,\bar{x}+ M.E.)$

Where,

Critical value:   $t_{(n-1),\alpha/2}=t_{(11-1),0.05/2}=t_{10,0.025}=2.228$ (From t-table)

By using MS-Excel Function "=TINV(0.05,10)")

$(22.64- 2.228*\frac{16.1386}{\sqrt{11}},22.64+2.228*\frac{16.1386}{\sqrt{11}})$

Margin of error=10.8414

The required C.I. for Age is