*Your* Random You Tube Videos

(5 pt)

Using your data set from the Random YouTube question, write down a 99% confidence interval for the proportion of YouTube videos that are not kid appropriate.

Be sure to use a t-distribution and the appropriate t^* multiplier!

from scipy.stats import t
from numpy import sqrt
p = 1
s = 0
tt = t.ppf(.99, df = 5)
[p-tt*s/sqrt(5), p+tt*s/sqrt(5)]

With a 99% confidence the proportion will fall within [1.0, 1.0] or 1.0±0 (based on my data set)

hat p = 3/5

T^ star = 4.60 (From T tables)

sigma = sqrt (((hat p* (1-hat p))/n) = 21.9 %

ME = T^star * sigma = 4.6*21.9 = 100 %

therefore n=5 is insufficient sample size to draw conclusions

For my 99% confidence interval:
t* = 4.60
hatp = 2/5
My calculation was (2/5) + or - 1.008

Are you trying to get us to prove a point about tiny data sets…?

In my YouTube sample, 3 videos were not child appropriate. So, \hat{p}=\frac{3}{5}, or 60% of videos aren’t appropriate for kids.

Assuming ME= \pm t^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} and a t^* value of 4.6 (from the tables), my margin of error simplifies to \pm \frac{4.6}{5} \sqrt{6} = \pm 2.254 … or, 60 \pm 225 % of videos on YouTube aren’t child appropriate.

All things considered, I suppose that’s both useless and probably true: some or all videos on YouTube might be inappropriate for children.

For my 99% confidence interval:

phat = 0

t* = 4.60

0 x 4.60 = 0

hat p = \frac{2}{5}

T^ast = 4.60

\sigma = \sqrt{(\frac{hat p (1- hat p)}{n})} = .219 or 21.9 %

\pm ME = (T^ast)(\sigma) = (4.60)(.219) = \pm1.008 or \pm 100.8 %

∴ n=5 is insufficient sample size to draw conclusions

For a 99% confidence interval of my data I first got…

p hat = 4/5

T*= 4.60

σ= 0.1789 or 17.89%

Then the + or - ME =(T*)(σ) = (4.60)(0.1789) which gets me + or - 0.8229 or 82.29%

p^ = 2.5/5
t* = 4.6
sigma = .22 or 22%
ME = (4.6)(.22) so… + or - 1.02 or 102%

p^ = 2/5
sigma=0.5477
t*=4.6041
ME=1.1278
[-0.7277683162333876, 1.5277683162333875]

For a 99% confidence interval

ˆp=1/5
T^*=4.6
σ=√(ˆp(1−ˆp)n) = .1789
±ME=(T^∗)(σ)=(4.60)(.1789)=±0.8229