One batch of walnut muffins uses

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

malfutka [58]

Answer:

99% confidence interval for the true mean number of words a third grader can read per minute is [35.5 , 35.9].

Step-by-step explanation:

We are given that a sample of 1584 third graders, the mean words per minute read was 35.7. Assume a population standard deviation of 3.3.

Firstly, the pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean words per minute read = 35.7

$\sigma$ = population standard deviation = 3.3

n = sample of third graders = 1584

$\mu$ = population mean number of words

Here for constructing 99% confidence interval we have used One-sample z test statistics as we know about the population standard deviation.

So, 99% confidence interval for the population mean, $\mu$ is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-2.58 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+2.58 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $35.7-2.58 \times {\frac{3.3}{\sqrt{1584} } }$ , $35.7+2.58 \times {\frac{3.3}{\sqrt{1584} } }$ ]

= [35.5 , 35.9]

Therefore, 99% confidence interval for the true mean number of words a third grader can read per minute is [35.5 , 35.9].

6 0

3 years ago

HELP! <3 I have other questions as well not yet posted, if you'd like more points! :3

ratelena [41]

Option B: $\frac{AB}{BC}=\frac{AE}{ED}$ is sufficient to prove that $BE$ ll $CD$

Explanation:

Given that ACD is a triangle.

The line EB intersects the sides AC and AD of the triangle.

The point E intersect the side AD and B intersect the side AC.

We need to prove that $BE$ ll $CD$

Then, the side splitter theorem states that "If a line intersects two sides of a triangle and divides the sides proportionally, the line is parallel to the third side of the triangle".

From the side splitter theorem, the line EB intersects the two sides of the triangle AC and AD and divides the sides proportionally.

Thus, the proportion of the sides is given by

$\frac{AB}{BC}=\frac{AE}{ED}$

This proportionality shows that the line is parallel to the third side of the triangle.

Hence, $\frac{AB}{BC}=\frac{AE}{ED}$ is sufficient to prove that $BE$ ll $CD$

Therefore, Option B is the correct answer.

3 0

3 years ago

Help me pls I really really need help with this I'm so bad at math

MAVERICK [17]

Question 7: Answer J (7/8 - 1/2 = 3/8)

6 0

3 years ago

5. A gardening club has 21 members, of which 13 are males and the rest are

Misha Larkins [42]

If you know that there are 21 members and 13 are females, if you subtract 13 from 21, you will get 8. There are 8 females in the gardening club.

Now you have to figure out the ratio to females to all club members. You already know that there are 8 females and 21 club members. The ratio would be 8:21.

I hope you found this helpful :)

3 0

3 years ago

Read 2 more answers

Help me please i need answers for my sisters homework

frozen [14]

Eleven Million Seven Hundred and Sixty Thousand Eight Hundred and Twenty-five.

4 0

3 years ago

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