A map has a scale of 1/2 inch = 20 miles.

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

I will give you the brainliest if you answer asap please

IRINA_888 [86]

Answer:

168

Step-by-step explanation:

You just need to triple the answer to get 168.

8 0

3 years ago

Convert 0.50 sen /g to RM/kg

IgorC [24]

the conversion is:

0.50 sen /g = 5 RM/kg

<h3>How to change the units?</h3>

First, we know that

1sen = 0.01RM

Then we can change:

0.50 sen /g = 0.50*(0.01 RM) /g = 0.005 RM/g

Now we use the relation:

1g = 0.001kg

So we can rewrite:

0.005 RM/g = 0.005 RM/(0.001kg) = 5 RM/kg

So the conversion is:

0.50 sen /g = 5 RM/kg

If you want to learn more about changes of units:

brainly.com/question/9032119

#SPJ1

7 0

2 years ago

Identify a counterexample to disprove 6n>n^2/6,where n is a real number.

scZoUnD [109]

Any number larger than 36 would be a counter examples.

When n is equal to 36, both sides of the inequality are equal to 216. As you get larger than 36 the right side is larger than the left side. Therefore, the inequality is not true.

3 0

3 years ago

54 1/4% equals _____

tresset_1 [31]

The answer to the question

8 0

3 years ago