# A random, data-based question on Confidence Intervals: 11:00 AM

edited February 18

(5 pts)

For this problem, you're going to compute a confidence interval for height based on some randomly generated data. Once you have your data and descriptive information, you'll compute your confidence interval in the form

$[bar{x} - z^** SE, bar{x} + z^** SE],$

where the $bar{x}$ is the mean of your sample and the standard error $SE = sigma//sqrt(n)$ is the standard deviation of your sample divided by the square root of the sample size. You should get your $z^**$-multiplier by using the calculator at the bottom of our normal calculator page.

You can get the specifics for your person question here.

## Comments

• edited February 17

My data has 98 rows. The mean of the heights is 67.90 and the standard deviation is 4.36. I'm going to use this information to compute a 92% confidence interval for height.

My standard error is:

$SE = sigma/sqrt(n) = 4.36/sqrt(98) = 0.4404.$

According to our normal calculator, my $z$-star multiplier is
$z^** =1.75$.

Thus, my margin of error is:

$ME = 1.75 times 0.4404 = 0.7707.$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME] = [67.9-0.7707, 67.9+0.7707] = [67.1293, 68.6707].$

• edited February 17

My data has 102 rows. The mean of my heights are 67.98 and the standard deviation is 4.42. Here is how I used this information to compute a 97% confidence interval for height.

I started by finding my standard error:

$.437645214 = 4.42//sqrt(102)$

I plugged the 97% percent into the calculator and got

$z^** = 2.17009$

Then I found my confidence level

$[67.98 - 2.17009 times .437645214 , 67.98 + 2.1009 times .437645214] = [67.0302705, 67.9297295]$

• edited February 17

Your data has 110 rows. The mean of your heights is 67.70 and the standard deviation is 4.03. Use this information to compute a 92% confidence interval for height. This is the information I used to compute a 92% confidence interval.

$SE = sigma/sqrt(n) = 4.03/sqrt(110) = 0.384245423465974$

to be 97% in the mean:
$z* = 1.750686$

then I found my confidence level:
[67.70 - 1.750686 x .384245423465974, 67.70 + 1.750686 x .384245423465974] = [67.04243562, 67.2536532)

• edited February 17

My data has 104 rows. The mean of your heights is 67.59 and the standard deviation is 4.73. Use this information to compute a 93% confidence interval for height.

My standard error is:
$SE= sigma/sqrt(n) = 4.73/sqrt(98) = 0.4778$

My $z^***$ multiplier is 1.812

My margin of error is

$ME =1.812 times 0.4778 = 0.8658$

Finally my confidence interval is...

[x¯−ME,x¯+ME]=[67.59−1.812,67.59+1.812]= [65.778, 69.402]

• edited February 17

My Data has 89 rows. The mean of the heights is 67.03 and a standard deviation of 4.38. I am going to use this information to compute a 91% confidence level for height. $SE = sigma/sqrt(n) = 4.38/sqrt(89)= 0.4627$

According to the normal caculator my margin of error is:
$ME= 175*0.4642= 0.81235$

Finally I found my confidence is:
$[x¯−ME,x¯+ME]=[67.03−0.8123,67.03+0.8123]= [66.2177, 67.8423]$

• edited February 17

There are 104 rows of data, the mean of my data is 67.06, and the standard deviation is 4.4.

The standard error is:

$SE = sigma/sqrt(n) = 4.4/sqrt(104) = .4315$

To be 97% confident in the mean:

$z^* = 2.1701$

All of this leads to a confidence interval of:

$[bar{x} - z^** SE, bar{x} + z^** SE] = [67.06 - 2.1701 * .4315 , 67.06 + 2.1701 * .4315] = [66.1236 , 67.9964]$

• edited February 17

My data has 90 rows. The mean of your heights is 66.90 and the standard deviation is 4.15. I will use this information to compute a 96% confidence interval for height.

I will start by finding my standard error:

$SE = sigma/sqrt(n) = 4.15/sqrt(90) =0.43744840965.$

To be 96% confident:

$z^* =2.053749.$

the margin of error is

$ME =2.053749 times 0.43744840965 =0.89840923387.$

Therefore my confidence interval is

$[bar(x) - ME, bar(x) + ME] = [66.9-2.053749, 66.9+2.053749] = [64.846251, 68.953749].$

• edited February 17

My data has 105 rows. The mean of heights is 66.99 and the standard deviation is 4.35. I'm going to use this information to compute a 98% confidence interval.

$SE = 4.35/sqrt(105) = .4245165$

According to our normal calculator, my $z$-star multiplier is

$z^*$ = .98757311

My margin of error is

$ME = 2.326348 * .42455165 = .9875311$

Then I found my confidence interval:

$[bar{x} - z^** SE, bar{x} + z^** SE]$

$(66.99 -( -2.326348 * .4245165))+(66.99 +(2.326348 * .4245165))$

• edited February 17

My data has 95 rows. The mean of your heights is 66.43 and the standard deviation is 4.04. Use this information to compute a 93% confidence interval for height.

My standard error is:

$SE = sigma/sqrt(n) = 4.04/sqrt(93) = 0.4144952542$

According to out normal calculator, my $z$-star multiplier is. $z^**=0.823814$. Thus, my margin of error is:

$ME=0.823814*0.4144952542=.3414669933$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME]=[66.43-.3414669933 , 66.43+.3414669933]=[66.08853300665649, 66.77146699334352]$

• edited February 17

My data has 93 rows. The mean of the heights is 66.5 and standard deviation is 4.49. I am going to use this information to compute a 95% confidence interval for height.

My standard error is:
$SE = sigma/sqrt(n) = 4.49/sqrt(93) = 0.46789432$

According to our normal calculator, my z-star multiplier is:

z*= 1.96

Thus my margin of error is:

1.96x .4679= .917084

Confidence interval:

[bar(x) - ME, bar(x) + ME]=[66.65-.917084 , 66.65+.917084]=[65.797008, 67.502992]

• edited February 18

Your data has 96 rows. The mean of the heights is 67.61 and the standard deviation is 4.64. I'm going to use this information to compute a 94% confidence interval for height.

My standard error is:

$SE= sigma/sqrt(n) = 4.64/sqrt(96) = 0.4735680169380811$

My $z$-star multiplier is
$z^** =1.880794$.
Thus my margin of error is:

$ME = 1.880794 times 0.4735680169380811 = 0.8906838848490414$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME] = [67.61- 0.890555959, 67.61+0.890555959] = (66.719444041, 68.500555959)$

• Here's some math:

$SE = sigma/sqrt(n) = 4.04/sqrt(93) = 0.4144952542$

• My data has 96 rows. The mean of your heights is 67.68 and the standard deviation is 4.32. I'm going to use this information to compute a 94% confidence interval for height.

My standard error is:

$SE = sigma/sqrt(n) = 4.32/sqrt(96) = 0.4409.$

According to our normal calculator, my $z$-star multiplier is
$z^** =1.88$.

Thus, my margin of error is:

$ME = 1.88 times 0.4409 = 0.8289.$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME] = [67.68-0.8289, 67.68+0.8289] = [66.8511, 68.5089].$

• My data has 99 rows. The mean of my heights is 66.55 and the standard deviation is 4.29. I'll use this information to compute a 92% confidence interval for height.

My standard error is:

$SE = sigma/sqrt(n) = 4.29/sqrt(99) = 0.4312.$

According to the normal calculator, my $z^**$ is 1.75. So my margin of error is:

$ME = 1.75 * 0.4312 = 0.7546$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME] = [66.55 - 0.7546, 66.55 + 0.7546] = [65.7954, 67.3046]$

• My data has 96 rows. The mean of my heights is 67.03, and the standard deviation is 4.71. We can use this information to compute a 93% confidence interval for height.

My standard error is

``````SE=σn−−√= 4.71/√96= .4807
``````

According to the normal calculator, my z-star multiplier is

``````z*= 0.93
``````

So, my margin of error is:

``````ME= .93 x .4807 = .447051
``````

Finally, my confidence interval is

``````[x¯−ME,x¯+ME] = [67.03-.447051, 67.03 +.447051] = [66.5829, 67.4771].
``````
• edited February 19

My data has 95 rows. The mean of your heights is 66.94 and the standard deviation is 4.31. Use this information to compute a 98% confidence interval for height

My standard error is:

$SE = sigma/sqrt(n) = 4.36/sqrt(95)=.4473$

According to our normal calculator, my $z^$ multiplier is $z^= 2.32$. Thus, my margin of error is :

$ME = 2.32 times 0.4473= 1.0377 Finally, my confidence interval is$[bar(x) - ME, bar(x) + ME] =[66.94-1.0377, 66.94+1.0377]= [65.9023, 67.9777]

• edited February 17

My data has 105 rows. The mean of my heights is 67.57 and the standard deviation is 4.18. I'm going to use this information to compute a 95% confidence interval for height.

My standard error is:

$SE = sigma//sqrt(n) = 4.18/sqrt(105) = .4079.$

According to our normal calculator, my z-star multiplier is

$z^** = 1.96$

Thus, my margin of error is:

$ME = 1.96 times 0.4079 = 1.5521.$

Finally, my confidence interval is:

$[bar(x) - ME, bar(x) + ME] = [67.57-1.5521, 67.57+1.5521] = [66.0179, 69.122].$

• My data has 92 rows. The mean of your heights is 66.23 and the standard deviation is 4.47. Use this information to compute a 90% confidence interval for height.

$SE = sigma/sqrt(n) = 4.47/sqrt(92) = 0.4660$

according to our normal calculator my $z$-star multiplier is
$z^**= 1.64$

Thus, my margin of error is:

$ME = 1.64 times 0.4660 = 0.7642.$

finally, my confidence interval is

$[bar(x) - ME, bar(x) + ME] = [66.23- 0.7642, 66.23 + 0.7642]= [65.4658, 66.9942] • edited February 17 My data has 108 rows. The mean of your heights is 66.88 and the standard deviation is 4.06. Use this information to compute a 92% confidence interval for height. SE = sigma/sqrt(n) = 4.06/sqrt(108) = .3906736822 according to our normal calculator my z-star multiplier is z*=1.75 Thus, my margin of error is: ME = 1.75 *.3906736822 = .6836789439 finally, my confidence interval is$[bar(x) - ME, bar(x) + ME] = [66.88-3906736822, 66.88+1.75] = [66.189, 67.571].

• My data has 110 rows. The mean of my heights is 67.13 and the standard deviation is 4.70. I'll use this information to compute a 96% confidence interval for height.

SE= sigma/sqrt(n)= 4.70/sqrt(110)=.448127417

According to our normal calculator my z* multiplier is z*=2.053749. So my margin of error is:

ME=2.053749* .448127417=.9203412346

Finally, my confidence interval is:
(bar(x) - ME, bar(x) + ME)= (67.13- .9203412346, 67.13 + .9203412346)= (66.20965877, 68.05034123)

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