Question

1. The table below gives the 20 values from the Class Survey: What Was Your High School G.P.A. taken in my afternoon class earlier this semester. At the time, we considered this data to be a sample, but for the purposes of this assignment, assume this data represents a population.

1.9

2.4

2.5

2.8

2.8

2.9

2.9

3.0

3.0

3.0

3.1

3.2

3.2

3.2

3.4

3.6

3.7

3.8

4.0

4.0

1. Use 1-Var Stats to calculate the mean, , and standard deviation, , of this population (round to two decimals). (2 pts)

2. Now, use the MATH – RANDINT function on your graphing calculator to choose an SRS of 8 values from this population (assign #s to all the data values, then choose 8 random integers from 1 – 28) and list them here. (3 pts)

3. Use 1-Var Stats to calculate the sample mean of the sample you chose in #2 (round to two decimals). (2 pts)

(continued on the back)

4. What is the critical value that should be used to construct a confidence interval with a confidence level of 90%. (3 pts)

5. Calculate the margin of error for constructing a 90% confidence interval estimate for the population mean (round to two decimals). (4 pts)

6. Calculate the upper and lower bounds for your confidence interval estimate. (4 pts)

7. Does your confidence interval contain the value of the population mean from #1?

​(It’s ok if it doesn’t; there should be a 10% chance that it will not.) (2 pts)

Answer 1

Given data is

1.9

2.4

2.5

2.8

2.8

2.9

2.9

3.0

3.0

3.0

3.1

3.2

3.2

3.2

3.4

3.6

3.7

3.8

4.0

4.0

a)To calculate the mean, , and standard deviation, , of this population

{Note - now you can use graphing calculator to calculate the mean, , and standard deviation , or it can be calculated by using mannual process or using any other software like R. Here i cannot paste output of using 1-Var Stats in graphing calculator so , will calculate it mannualy and then using R-software .}

Formula -        Mean = =

           standard deviation =

Hence Mean of population is

= = ( 1.9 + 2.4 + 2.5+ 2.8 +2.8 +2.9 +......+3.4+ 3.6 +3.7 +3.8+ 4.0+4.0) / 20

     = 62.4 / 20

= 3.12                  ( Population mean )

And Standard Deviation is

=

Now =   { (1.9-3.12)2 + (2.4-3.12)2 + (2.5-3.12)2+ (2.8-3.12)2 +......... +(3.7-3.12)2 +(3.8-3.12)2+

                                           (4.0-3.12)2+(4.0-3.12)2)   }/ (20-1)

         = 5.412 / 19 = 0.2848421

Hence = = = 0.533706

Thus mean, , and standard deviation, of this population (round to two decimals) is

Mean = 3.12

Standard Deviation = 0.53

2. Now, use the MATH – RANDINT function on your graphing calculator to choose an SRS of 8 values from this population

Note You can genrate random numbers using graphing calculator( by using MATH – RANDINT) or by statistical book. Here i cannot paste output of graphing calculator hence genrate random nubers using R-software only .

R - Code and command

> r=1:20                      # define set of numbers
> R_N=sample(r,8)    # to genrate 8 random numbers between 1-20
> R_N                            # these are genrate random numbers
[1] 7 6 14 16 9 5 13 4

# Now we will import population data into R

> x=scan("clipboard")
Read 20 items
> x                                    # given population data
[1] 1.9 2.4 2.5 2.8 2.8 2.9 2.9 3.0 3.0 3.0 3.1 3.2 3.2 3.2 3.4 3.6 3.7 3.8 4.0
[20] 4.0

> Sample=x[R_N]           # Data corespong to genrated Random number

> data.frame("Random Number"=R_N,Sample)

Random.Number   Sample
1              7                     2.9
2              6                     2.9
3            14                     3.2
4            16   3.6
5            9                     3.0
6              5       2.8
7            13       3.2
8              4    2.8

So these are SRS of size 8 :- 2.9 , 2.9 , 3.2 , 3.6 , 3.0 , 2.8 , 3.2 , 2.8

3. Use 1-Var Stats to calculate the sample mean of the sample you chose in #2

= = ( 2.9+2.9+ 3.2 + 3.6 + 3.0 + 2.8 + 3.2 + 2.8) / 8

       = 24.4 / 8 = 3.05

= 3.05                 ( Sample mean )

And Standard Deviation of sample is

=

Now =   { (2.9-3.05 )2 + (2.9-3.05 )2 + (3.2-3.05 )2+ (3.6-3.12)2 +.........

                                                  +(3.2-3.05 )2 +(2.8-3.05 )2 } / ( 8-1 )

         = 0.52 / 7 = 0.07428571

Hence = = = 0.2725541

Thus mean, , and standard deviation, of this Sample (round to two decimals) is

Mean = 3.05

Standard Deviation = 0.27

4. What is the critical value that should be used to construct a confidence interval with a confidence level of 90%.

For confidence level of 90%. level of significance will be = 0.10 thus

Critical value is = 1.64              ( at = 1.64 , = 0.05 }

Hence Critical value is 1.64

                                        

5. Calculate the margin of error for constructing a 90% confidence interval estimate for the population mean (round to two decimals)

Formula -

Margin of error M.E =   * /

Here = 1.64   , = 0.2725541 and n = 8 ( sample size)

Thus M.E =   * / = 1.64 * 0.2725541 / 81/2

                                                   = 0.1580344

Thus Margin of error for constructing a 90% confidence interval estimate for the population mean (round to two decimals) is 0.16

6. Calculate the upper and lower bounds for your confidence interval estimate.

Upper bounds for confidence interval is

U.B = + M.E = 3.05 + 0.16 = 3.21

Lower bounds for confidence interval is

L.B = - M.E = 3.05 - 0.16 = 2.89

Hence upper and lower bounds for your confidence interval estimate are 3.21 and 2.89 respectively

I.e 90% Confidence interval is ( 2.89 , 3.21 )

7. Does your confidence interval contain the value of the population mean from #1

Our populatio means is 3.12 ( from #1)

Hence confidence interval ( 2.89 , 3.21 ) contain the value of the population mean 3.12 from #1