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jeyben [28]
3 years ago
9

There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on

good measures of the outcome. One frequently used test is the DRP or Degree of Reading Power. A researcher suspects that the mean score µ of all third graders in Henrico County Schools is different from the national mean, which is 32. To test her suspicion, she administers the DRP to an SRS of 44 Henrico County third-grade students and the standard deviation of scores in Henrico County Schools is known to be σ = 11 Their scores were: 40 26 39 14 42 18 25 43 46 27 19 47 19 26 35 34 15 44 40 38 31 46 52 25 35 35 33 29 34 41 49 28 42 47 35 48 22 33 41 51 27 14 54 45
a.Construct a 90% confidence interval to estimate the mean DRP score in Henrico County Schools.



b.Use the confidence interval you constructed in part (a) to comment on whether you agree with the researcher’s claim. Explain your reasoning clearly.
Mathematics
1 answer:
ELEN [110]3 years ago
6 0

Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools  is =32±2.7280

i.e. C.I.  [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

True mean =32

Standard deviation=11

And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

C.I.=mean±Z[Standard  deviation/Sqrt(n)]

=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

=32±2.7280

i.e. C.I.  [29.3,34.7]

Now Here to commit on calculated interval we should check  value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

Z-score=(sample mean -true mean)/[standard deviation/Sqrt(n)]

=(34.86-32)/[11/Sqrt(44)]

Z=1.72465

For ,

P(Z≥1.72465)=0.08544

For 0.05 significance level  the results are not significant at  p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

Hence the results are not significant.

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The solutions of a quadratic equation are ​-4,9 Write a related quadratic function in factored form.
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Answer:

y = (x + 4)(x - 9)

Step-by-step explanation:

x = {-4, 9}

y = (x + 4)(x - 9)

3 0
3 years ago
Julie spends $5.62 at the store. Micah spends 5 times as much as Julie. Jeremy spends $6.72 more than Micah. How much money does
avanturin [10]
Micah spent 28.10 dollars, Jeremy spent 34.82 dollars
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3 years ago
Can you help me find each measure?
madam [21]

Answer:

Step-by-step explanation:

1. the measure for HJ you can already see, start from h and draw a line to J. The measure shows 63 degrees.

2. start at F and draw a line to G first. the measure shows 65 degrees. Now continue the line from where you stopped at G, to H. There's no measure but you can see it is in a semicircle. A semicircle is 180 degrees andthe other two angles are 63 and 65.

So...

180=63+65+GH

subtract to get GH

180-63-65= 52

so the measure GH is 52 degrees. The full measure you are trying to find is FGH thought. So ad the 65 and the 52.

FGH =117 degrees.

3. The meaure is CDE. CD as you can see is a right angles, so 90 degrees. But there is no measure for DE. If you look to the angle vertical from DE which is BA. It measures 40 degrees. DE and BA are vertical so they are congruent. If DE equals 40  and CD equals 90, put them together and you get 130.

CDE= 130 degrees

4. Next measure is BCD. We already know CD is a 90 degree angle but BC is blank. You can see the measure BCD is in a semi circle. A semicircle equals 180, CD equals 90, and BA equals 40

so...

180= 90+40+BC

so subtract

180-90-40= 50

BC =50

so add BC=50 and CD=90. SO, BCD is 140 degrees.

5.The angle LMN is next. MN is 30 but LM is blank. LMN is in a semi circle.

A semicircle is 180 degrees.

so...

180=105+30+LM

subtract

180-105-30=45

LM = 45

Add LM=45 and MN=30

LMN is 75 degrees.

6.The last angle is LNP

If you look at it MP is a semi circle, so 180 degrees. And LM is 45 from our last question. so 180 +45=225

LNP=225

hope this helps

4 0
3 years ago
What is the distance between 178 and 313 on a number line?
Damm [24]

Answer:

135 is the distance

Step-by-step explanation:

I subtracted the numbers.

I hope this helped and have a great day!

(please vote this brainliest)

4 0
3 years ago
Read 2 more answers
Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.
DENIUS [597]

Answer:

The probability that a randomly chosen plate contains the number 2222 is 0.000028 approximately.

The probability that a randomly chosen plate contains the sub-string HI is 0.002548  approximately.

Step-by-step explanation:

Consider the provided information.

Illinois license plates used to consist of either three letters followed by three digits or two letters followed by four digits.

Part (A)

Let A is the ways in which plates consist of three letters followed by three digits and B is the ways in which two letters followed by four digits.

Here repetition is allow. The number of alphabets are 26 and the number of distinct digits are 10.

The numbers of ways in which three letters followed by three digits can be chosen is: 26\times 26\times 26 \times 10 \times 10 \times10

26^3\times 10^3=17576000

The numbers of ways in which two letters followed by four digits can be chosen is: 26\times 26\times 10 \times 10 \times 10 \times10

26^2\times 10^4=6760000

Hence, the total number of ways are 17576000 + 6760000 = 24336000

Randomly chosen plate contains the number 2222 that means the first two letter can be any alphabets but the rest of the digit should be 2222.

Thus, the total number of ways that a randomly chosen plate contains the number 2222 number are: 26^2=676

The probability that a randomly chosen plate contains the number 2222 is:

\frac{676}{24336000} \approx 0.000028

Part (B)

The number of ways in which chosen plate contains the sub-string HI:

If three letters followed by three digits plate contains the sub-string HI, then the number of possible ways are:

26\times 1\times10^3+1\times 26\times10^3

If two letters followed by four digits plate contains the sub-string HI, then the number of possible ways are:

1\times 10^4

Thus, the total number of ways that a randomly chosen plate contains the sub-string HI are:

1\times 10^4+26\times 1\times10^3+1\times 26\times10^3

62000

From part (A) we know that the total number of ways to chose a number plate is 24336000.

The probability that a randomly chosen plate contains the sub-string HI is:

\frac{62000}{24336000} \approx 0.002548

7 0
3 years ago
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