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A polling company conducted a survey asking 40 fourth year students from Ohio and 33 fourth-year students from Pennsylvania abou

miv72 [106K]

Answer:

The 99.9% confidence interval for the difference between these results is between 12.09 and 25.91 hours per week.

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction of normal variables:

When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of variances.

Ohio had an average of 52, standard deviation of 7.22, sample of 40.

This means that $\mu_O = 52, s_O = \frac{7.22}{\sqrt{40}} = 1.1416$

Pennsylvania had an average of 33, standard deviation of 10.11, sample of 33.

This means that $\mu_P = 33, s_P = \frac{10.11}{\sqrt{33}} = 1.76$

Distribution of the difference:

$\mu = \mu_O - \mu_P = 52 - 33 = 19$

$s = \sqrt{s_O^2 + s_P^2} = \sqrt{1.1416^2 + 1.76^2} = 2.1$

Confidence interval:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.999}{2} = 0.0005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.0005 = 0.9995$ , so Z = 3.29.

Now, find the margin of error M as such

$M = zs$

$M = 3.29*2.1 = 6.91$

The lower end of the interval is the sample mean subtracted by M. So it is 19 - 6.91 = 12.09 hours per week

The upper end of the interval is the sample mean added to M. So it is 19 + 6.91 = 25.91 hours per week.

The 99.9% confidence interval for the difference between these results is between 12.09 and 25.91 hours per week.

8 0

3 years ago

The cost of renting a sailboat at a lake is $70

zmey [24]

Answer:

y=70x+12

Step-by-step explanation:

y= amount of total money spent

x= amount of hours

6 0

3 years ago

55 men can finish a piece of work in 42 days. How many additional men must be added to complete the work 9 days earlier.

Alisiya [41]

$55 . 42 = x . 33\\55 . 14 = 11 . x\\x = 5 . 14 = 70\\70 - 55 = 15$ Answer:

Step-by-step explanation:

6 0

4 years ago

A number is chosen at random from 1 to 10. Find the probability of not selecting

PilotLPTM [1.2K]

Answer:

0.3

Step-by-step explanation:

Number of ways of selecting r elements from a set of n different elements is given by

C(n,r)=(n r)=n!/(r!(n−r)!)

No of ways of selecting one number out of ten numbers from one to 10 is 10

It can also be calculated using

C(n,r)=(n r)=n!/(r!(n−r)!)

where n = 10 and r = 1

C(10,1)=(10 1)=10!(1!(10−1)!) = 10*9!/ 1!*9! = 10

multiples of 2 in range 1 to 10 are 2, 4, 6, 8 , 10

multiples of 3 in range 1 to 10 are 3, 6, 9

Therefore number which are multiple of 2 and 3 are

2, 3,4,6,8,9,10 ( 7 numbers)

therefore no of ways of selecting multiple of 2 and 3 is 7 ways

number which are not multiple of 2 and 3 in range 1 to 10

1,5,7 (3 numbers)

no of ways of not selecting multiple of 2 and 3 is 3 ways

It can also be calculated using

C(3,1)=(3 1)=3!/(1!(3−1)!)

where n = 3 and r = 1

C(3,1)=(3 1)=3!/(1!(3−1)!) = 3!/(1!(2)!)= 3*2!/ 1!*2! = 2

Therefore no of ways of not selecting multiple of 2 and 3 is 3 ways

probability of not selecting a multiple of 2 or a multiple of 3 = no of ways of not selecting multiple of 2 and 3 /No of ways of selecting one number out of one to 10 = 3/10 = 0.3

4 0

3 years ago

Someone please help :( what is the slope??

statuscvo [17]

Answer:

Step-by-step explanation:

look up a slope calculator online

6 0

3 years ago

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