What is less than 1/6 then 6/6 ? it will help me a lot plz

You recently sent out a survey to determine if the percentage of adults who use social media has changed from 66%, which was the

il63 [147K]

Answer:

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 2809 people who responded to survey, 1634 stated that they currently use social media.

This means that $n = 2809, \pi = \frac{1634}{2809} = 0.5817$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a pvalue of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 - 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.56$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5817 + 2.327\sqrt{\frac{0.5817*4183}{2809}} = 0.6034$

The 98% confidence interval estimate of the proportion of adults who use social media is (0.56, 0.6034).

6 0

2 years ago

WILL MARK YOU BRAINLIEST HELP ASAP

Marat540 [252]

Answer:

I believe it is c> 34

Step-by-step explanation:

Correct me if i'm wrong

6 0

2 years ago

Read 2 more answers

the cost of a school banquet is $95 plus $15 for each person attending. write an equation that gives total cost as a function of

AleksandrR [38]

Im sorry if I misunderstood the problem but heres my take.

Answer:

1155 for the cost of the people and 1250 for the total including the $95

Step-by-step explanation:

Multiply it ( 77 x 15 )

then add the answer to the $95 ( 95 + N )

= $N

8 0

2 years ago

Suppose in the University of Manitoba, 40% of the students live in apartments. If 600 students are randomly selected, then the a

just olya [345]

Answer:

$P(200\leq X\leq 400)=P(\frac{200-240}{12}\leq \frac{X-\mu}{\sigma}\leq \frac{400-240}{12})=P(-3.33\leq Z \leq 13.33)$

$=P(Z$

So then the correct answer would be:

B) .9996

Step-by-step explanation:

The exact way to solve this problem is using the binomial distribution, assuming that our random variable of interest is "number of students living in apartments" represented by X and $X \sim Bin(n=600, p =0.4)$

And we want this probability:

$P(200 \leq X \leq 400) [tex]In order to find this probability we can use the foloowing excel code:"=BINOM.DIST(400,600,0.4,TRUE)-BINOM.DIST(199,600,0.4,TRUE)"And we got:[tex] P(200 \leq X \leq 400) =0.999675[tex]But for this case the problem says that we need to approximate, so then we can use the normal approximation to the normal distribution. We need to check the conditions in order to use the normal approximation.
[tex]np=600*0.4=240\geq 10$