Confused on number 10. I put one solution, but I don’t know if it’s right. I got

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

last month , he sold 50 chickens and 30 ducks for $550 . This month , he sold 44 chickens and 36 ducks for $532. How much does a

Kazeer [188]

Hello there, my fellow human being!
So, the chicken costs $8 and the duck costs $5, here's why.

Let's say x is the cost of a chicken while y is the cost of a duck.
We can make two linear equations using the information above.
Last month, he sold 50 chickens and 30 ducks for $550: 50x + 30y= 550
This month, he sold 44 chicken and 36 ducks for $532: 44x + 36y = 532

50x/10 + 30y/10 = 440/10
44x/4 + 36y/4 = 532/4

5x + 3y = 44
11x + 9y = 133

So, now that we have our answer simplified, we have to use elimination to solve this system of equations. But, first we need to make sure that at least one of our variables is able to be canceled out.
Let's multiply this equation by -3.
(5x + 3y = 44) * -3.

-15x-9y=-165.

11x + 9y = 133
-15x - 9y = -165
______________
-4x/4 = -32/4
x = 8
11x + 9y = 133
11(8) + 9y = 133
88 + 9y = 133
-88 -88
________________
9y/9 = 45/9
y = 5.

4 0

3 years ago

Read 2 more answers

Is the sequence geometric? if so, identify the common.<br> 2, 6, 18, 54,...

Nastasia [14]

Common ratio = 6/2 = 3

7 0

3 years ago

Read 2 more answers

Paige launched a ball using a catapult she built. The height of the ball (in meters above the ground) ttt seconds after launch i

agasfer [191]

Answer: