Find the percent change from 72 to 80.

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

Part b

If we replace the values obtained we got:

$0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429$

$0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491$

The 99% confidence interval would be given by (0.429;0.491)

8 0

3 years ago

Look at the 2 fractions below. Choose which are equivalent fractions from the choices. Hint: There may be more than 1 correct an

sergeinik [125]

2/12 and 9/12 is right. 4/24 and 18/24 is right. So, answers A and D are right

6 0

3 years ago

Read 2 more answers

Help if you do this you will get seven years of good luck if you don't you will get seven years of bad luck: (SAT Prep): Find th

KatRina [158]

Answer:

$Slope=\frac{7}{2}=3.5$

Step-by-step explanation:

Slope of line: $Slope\ of\ line\ passing\ through\ (x_1,y_1)\ and\ (x_2,y_2)\ is\ given\ by\ \frac{y_2-y_1}{x_2-x_1}$

$Here\ (x_1,y_1)=(\frac{2}{7},5)\\\\(x_2,y_2)=(10,39)\\\\Slope=\frac{39-5}{10-\frac{2}{7}}\\\\Slope=\frac{34}{\frac{68}{7}}\\\\Slope=\frac{34\times 7}{68}\\\\Slope=\frac{7}{2}=3.5$

5 0

3 years ago

HELP please urgent!! - PLEASE CLICK, NEED HELP -

Makovka662 [10]

Answer:

$y=3x^2-12x-135$

Step-by-step explanation:

The standard form of a quadratic is $y=ax^2+bx+c$

We will use the x and y values from each of our 3 points to find a, b, and c. Filling in the x and y values from each point:

First point (-5, 0):

$0=a(-5)^2+b(-5)+c$ and

0 = 25a - 5b + c

Second point (9, 0):

$0=a(9)^2+b(9)+c$ and

0 = 81a + 9b + c

Third point (8, -39):

$-39=a(8)^2+b(8)+c$ and

-39 = 64a + 8b + c

Use the elimination method of solving systems on the first 2 equations to eliminate the c. Multiply the first equation by -1 to get:

-25a + 5b - c = 0

81a + 9b + c = 0

When the c's cancel out you're left with

56a + 14b = 0

Now use the second and third equations and elimination to get rid of the c's. Multiply the second equation by -1 to get:

-81a - 9b - c = 0

64a + 8b + c = -39

When the c's cancel out you're left with

-17a - 1b = -39

Between those 2 bolded equations, eliminate the b's. Do this by multiplying the second of the 2 by 14 to get:

56a + 14b = 0

-238a - 14b = -546

When the b's cancel out you're left with

-182a = -546 and

a = 3

Use this value of a to back substitute to find b:

56a + 14b = 0 so 56(3) + 14b = 0 gives you

168 + 14b = 0 and 14b = -168 so

b = -12

Now back sub in a and b to find c:

0 = 25a - 5b + c gives you

0 = 75+ 60 + c so

0 = 135 + c and

c = -135

Put that all together into the standard form equation to get

$y=3x^2-12x-135$

6 0

4 years ago

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Hi guys my little brother needs help to Do his homework And I’m doing mine so I can’t help him can anyone help plzz

alukav5142 [94]

The answer is 91.77

22.89 x 2 =45.78

45.78 + 45.99 = 91.77

3 0

4 years ago