Ask question

Ask question

All categories

3 years ago

7

Ay anyone willing to help me with my work kinda lost

1 answer:

kirill115 [55]3 years ago

7 0

Okaayyy

Set B has the greater mode

You might be interested in

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B200%5C%5C" id="TexFormula1" title="\sqrt{200\\" alt="\sqrt{200\\" align="absmiddle"

Yuri [45]

Do sqrt(100) * sqrt(2)
Which will be 10 * sqrt(2)

4 0

2 years ago

Read 2 more answers

Derrick has $4 to play video games at the mall each game cost a quarter to play. What’s the total number of games he can play

Alborosie

Derrick can play 16 games in total.

5 0

2 years ago

Read 2 more answers

How do you solve the equation 2/7m-1/7=3/14

photoshop1234 [79]

Add 1/7 to both sides to get 2/7m=3/14+1/7 find common denominater (14) so multipy 1/7 by 2 to get 2/14 than add that to 3/14 to get 5/14 than take 2/7 and 5/14 turn into a decimals than divide the 2/7 on both sides to get m=?

8 0

3 years ago

Mr. Faber has 84 pear trees and 180 apple trees. He wants to plant the trees in rows of equal width. What is the largest amount

Galina-37 [17]

I believe it is 15. Hope that helps!

8 0

2 years ago

Read 2 more answers