# Confidence intervals and sample size responses

INSTRUCTIONS: Provide (2) 150 words response for RESPONSES 1 AND 2 below. Responses may include direct questions. In your first peer posts, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine. Have fun and be creative with it and calculate another T-confidence interval and interpret your results. Compare your results to that of the initial 95%, how much do they differ? How useful can this type of information be when you go to buy a new car, or even a house?

In your second peer posts, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine. Have fun and be creative with it and calculate another proportion interval and interpret your results. Compare your results to that of the initial 95%, how much do they differ? How useful can this type of information be when you go to buy a new car, or even a house?

Attached are the excel docs for both responses to help with the post.

RESPONSE 1:

FIRST

• To      calculate for 95% confidence

1-0.95=0.05

• Calculate      the confidence interval (NEEDS TO BE DIVIDED BY 2)

0.05/2= 0.025

• ***IMPORTANT***      (I missed this the first calculation around, you will get a very wrong      negative number if you do not remember the more than L)

-Change to less than equation for Excel function

1-0.025= 0.975. α =0.975

• Calculate      degrees of freedom (DF)

Take N-1

10-1=9

• Calculate      for T-critical value:

=T.INV(0.975,9) = 2.26

T-critical value = 2.26

• At      this point, we have everything we need to plug it into excel – YAY!

????̅= Mean (40,589)

????= .975

N-1= 9

????????= 15768.88432

N= 10

T-crit= 2.26

Confidence Level @ 95%= 11280.3803 (Use your toolpack to get this number) OR wait for a detailed explanation below.

• FINALLY      (use the given equation from given pdf).

2.26 *(15768.88432/SQRT(10))= 11280.3803

=40,589± 11280.3803

(29308, 51869). Thus, I am 95% confident that the population mean car price for the type of cars I selected in Wk 1 is between \$29,308 and \$51,869. I realize this is quite the disparity, but I believe this is attributed to the wide range of cars I chose.

SECOND

• calculate      the success rate (p) and failure (q)

P=0.7

Q=0.3

• Calculate      Z critical value of 95% confidence

α =0.975

=NORM.S.INV(0.975)= 1.96

Z critical value=1.96

• Now,      you have everything necessary to plug into the given equation. Go for it!

.70 ± 1.96(.1449137)

0.70 ± 0.28403=

(.41597, .98403). Thus, I am 95% confident that the population proportion of car prices that are less than the mean is between 41.6% and 98.4%. Please let me know if you got a different calculation here. I was confused as to why it was so similar to the example, when part 1 was not. Have a great week!

RESPONSE 2:

Step 1.

We are asked to determine the T- confidence interval for a sample of car population prices.

????̅± ???? ∗ ( ???????? √???? )

Step 2.

To calculate for 95% confidence

1-0.95=0.05

Step 3.

Calculate the confidence interval (NEEDS TO BE DIVIDED BY 2)

0.05/2= 0.025

Step 4: (IMPORTANT)

Change to less than equation for Excel function

1-0.025= 0.975. α =0.975

Step 5:

Calculate degrees of freedom (DF)

-Take N -1

10-1=9

Everything we have for Excel function!

• ????̅= Mean (56,813.70)
• ????=      T.INV( .975, 9)
• ????????=      33061.48742
• N= 10

Excel Function:

=T.INV(0.975,9) = 2.26

T-critical value- 2.26

Equation should look like:

=2.26 *(33061.48742/SQRT(10))= 23650.76334

=56,813.70 ± 23650.76334

Equals (33,162.94; 80,464.46)

I am 95% confidence that the sample price of the cars will be between \$33,162.94 and \$80,464.46

Part 2:

Calculate the proportion confidence interval with the proportion of the number of vehicles that fall below the average.

????̂± ???? ∗ (√ ????̂∗????̂/ ???? )

Step 1:

calculate the success rate (p) and failure (q)

P=0.6

Q=0.4

Step 2:

Calculate Z critical value of 95% confidence

α =0.975

=NORM.S.INV(0.975)= 1.96

Z critical value=1.96

Step 3:

Put values in function

1.96 * ( √(.6*.4)/10)

0.6 ± 0.3036

(0.2963,0.9036)

I am 95% confident that the proportion of cars sampled that will fall below the average goes from 30% to 90%

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