What is the coefficient in the term k?<br> A.0<br> B.1<br> C.2<br> D.k

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

2 years ago

HELPPPPPPPPPPPPPPPPPPPPP

77julia77 [94]

Answer:

ill guess -2 for u

Step-by-step explanation:

5 0

1 year ago

Complete the division problem by determining the number that should be placed in the box.

Assoli18 [71]

The solution 349 is given, and its number 3, 4 and 9 determine how many times 13 can be subtracted:

The first subtraction is 3x13 (x100) = 3900
The second subtraction is 4x13 (x10) = 520
The third subtraction is 6x13 = 78

So the answer is 520, which is also what you get when you subtract 78 from 598 of course.

6 0

3 years ago

Read 2 more answers

Question 5 (1 point)

Leokris [45]

A: Add up to 180 degrees

8 0

2 years ago

Suppose that f(x)=x^2 and g(x)=-1/4(x+7)^2. Which statement best compares the graph of g(x) with the graph of f(x)?

wolverine [178]

Answer:

Both of these graphs are quadratics. They both form a parabola opening up

Step-by-step explanation:

The only difference between g(x) and f(x) is that g(x) is moved to the left 7 unit and has a horizontal stretch of 1/4

5 0

2 years ago

What is the coefficient in the term k? A.0 B.1 C.2 D.k