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 May 13th, 2017, 06:29 PM #1 Newbie   Joined: May 2017 From: Texas Posts: 2 Thanks: 0 I need urgent Stats help on two questions. I've barely got a grasp on my HS Stats class and its getting to that end of the year hurtle and i'm catching up on missed lessons while I was away. I will give many thanks to the kind one to lend a hand. 1) Check to see if the following sample sizes are large enough to be approximated by a normal curve? a) n = 50 & p = .3 e) n = 50 & p = .05 b) n = 15 & p = .45 f) n = 100 & p = .01 c) n = 100 & p = .7 g) n = 40 & p = .25 d) n = 60 & p = .25 h) n = 80 & p = .9 2) USA Today reported that 36% of adult drivers admit that they often or sometimes talk on a cell phone when driving. This was based on a random sample of 1004 adult drivers. a) What is a 90% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving? b) What is a 95% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving? c) What is a 99% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving? d) What do you notice? Last edited by kaid; May 13th, 2017 at 07:19 PM.
May 14th, 2017, 02:37 PM   #2
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Quote:
 Originally Posted by kaid I've barely got a grasp on my HS Stats class and its getting to that end of the year hurtle and i'm catching up on missed lessons while I was away. I will give many thanks to the kind one to lend a hand. 1) Check to see if the following sample sizes are large enough to be approximated by a normal curve? a) n = 50 & p = .3 e) n = 50 & p = .05 b) n = 15 & p = .45 f) n = 100 & p = .01 c) n = 100 & p = .7 g) n = 40 & p = .25 d) n = 60 & p = .25 h) n = 80 & p = .9
using the normal approximation is appropriate when $n p \geq 5$

Quote:
 2) USA Today reported that 36% of adult drivers admit that they often or sometimes talk on a cell phone when driving. This was based on a random sample of 1004 adult drivers. a) What is a 90% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving?
$p=0.36$

$\sigma = \dfrac{p(1-p)}{\sqrt{1004}} \approx 0.00727$

a $90\%$ significance corresponds to a z-score of

$z = \Phi^{-1}\left(\dfrac{1-0.9}{2}\right) \approx -1.64485$

The $90\%$ confidence interval is thus

$I_{0.9}=[0.36 - |z| \sigma, ~0.36 + |z|\sigma] \approx [0.34804,0.37196]$

Quote:
 b) What is a 95% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving? c) What is a 99% confidence interval for the true proportion of adult drivers who have often or sometimes talk on a cell phone when driving? d) What do you notice?
just repeat the calcs above finding the appropriate z-score for each significance level.

what you observe should be intuitively clear.

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