What is the y-intercept of the graph of x=9y-5

According to a Pew Research Center, in May 2011, 35% of all American adults had a smart phone (one which the user can use to rea

Veronika [31]

Answer:

$p_v =P(z>1.82)=0.034$

Assuming a standard significance level of $\alpha=0.05$ the best conclusion for this case would be:

4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).

Because $p_v$

If we select a significance level lower than 0.034 then the conclusion would change.

Step-by-step explanation:

Data given

n=300 represent the random sample taken

X=120 represent the people who have a smart phone

$\hat p=\frac{120}{300}=0.4$ estimated proportion of people who have a smart phone

$p_o=0.35$ is the value that we want to test

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion of people who have a smart phone is higher than 0.35, the system of hypothesis are.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

Statistical decision

Since is a right tailed test the p value would be:

$p_v =P(z>1.82)=0.034$

Assuming a standard significance level of $\alpha=0.05$ the best conclusion for this case would be:

4. There is not enough evidence to show that more than 35% of community college students own a smart phone (P-value = 0.034).

Because $p_v$

If we select a significance level lower than 0.034 then the conclusion would change.

4 0

3 years ago

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r

svp [43]

Answer:

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that $a = 50, b = 52$

If one such class is randomly selected, find the probability that the class length is between 51.5 and 51.7 min.

$P(51.5 \leq X \leq 51.7) = \frac{51.7 - 51.5}{52 - 50} = \frac{0.2}{2} = 0.1$

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

3 0

3 years ago

What is 0.25 (repeating 5) as a fraction?

Vanyuwa [196]

1/2 or 2/4 pretty sure

4 0

3 years ago

Please help Fast A parking lot contains motorcycles (2 Wheels) and cars (4 Wheels). There are 35 vehicles and 114 wheels. How ma

swat32

Answer: There will be 20 motorcycles and 26 cars

let me know if you want me to explain:)

Step-by-step explanation:

If you want to check multiply 26 by the amount of wheels on each car (4) to get 104 total car wheels, then multiply the 20 motorcycles by the wheels of each motorcycle (2) to get 40. Lastly add the 104 and 40 together to get 144 the total amount of wheels in the parking lot.

Hope this helps :)

3 0

3 years ago

The smiths bought an apartment for $75,000. assuming that the value of the apartment will appreciate at most 4% a year, how many

DedPeter [7]

Answer:

1st year is $75,000 * 0.04 = $3,000 $75,000+$3,000=$78,000

Step-by-step explanation:

6 0

3 years ago