Your class is learning how to tie knots.each student needs a peace of rope thatis 3/8 yards. how much do we need for 16 stdents

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

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

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

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

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

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

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

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

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

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

3 years ago

A recipe for chilie uses 2 1/3 cups of beans. Keith is making 1 1/2 times the recipe. Which equation helps keith find how many c

yuradex [85]

Answer:

$2\frac{1}{3} * 1\frac{1}{2} = 3\frac{1}{2}$

Step-by-step explanation:

Given that a recipe of chilie requires $2\frac{1}{3}$ cups of beans, to make $1\frac{1}{2}$ times the recipe, Keith would need x cups of beans.

Thus:

1 recipe = $2\frac{1}{3}$ cups of beans

$1\frac{1}{2}$ of the recipe = x cups of beans

$x*1 = 2\frac{1}{3} * 1\frac{1}{2}$

$x = \frac{7}{3} * \frac{3}{2}$

$x = \frac{7}{2}$

$x = 3\frac{1}{2}$

Therefore the equation that helps Keith find how many cups of beans he needs is: $2\frac{1}{3} * 1\frac{1}{2} = 3\frac{1}{2}$

6 0

2 years ago

What is the answer because I don’t understand

Ainat [17]

Answer: 12a-a (hope this helps :)

=12a-1a

=12a+-1a

=a+a+a+a+a+a+a+a+

=11A

7 0

2 years ago

PLEASE Which of the following are correct statements about the slope formula?

liq [111]

Slope formula is:

(y₂ - y₁)/(x₂ - x₁) = slope, or (C) (numbering varied)

The other slope formula:

y = rise (as it controls the up and down), while x = run (controls left to right).

y is over x, or rise/run (D)

hope this helps

3 0

2 years ago

Read 2 more answers

HELP ME PLEASE PLEASE HURRY MATH FOR A P EX

Murljashka [212]

D
where they meet at is 2,6

7 0

3 years ago

Read 2 more answers