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andriy [413]
3 years ago
14

A simple random sample of 20 third-grade children from a certain school district is selected, and each is given a test to measur

e his/her reading ability. You are interested in calculating a 95% confidence interval for the population mean score. In the sample, the mean score is 64 points, and the standard deviation is 12 points.
What is the margin of error associated with the confidence interval?

A. 2.68 points

B. 4.64 points

C. 5.62 points

D. 6.84 point
Mathematics
1 answer:
ICE Princess25 [194]3 years ago
4 0

Answer:

Option C) 5.62 points  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 20

Sample mean score = 64

Sample standard deviation, s = 12

Degree of freedom =

=n-1\\=20-1\\=19

We have to calculate margin of error for a 95% confidence interval.

Formula for margin of error:

t_{critical}\times \displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.05} = \pm 2.093  

2.093\times (\displaystyle\frac{12}{\sqrt{20}} )\\\\ =5.6161\approx 5.62  

Thus, the correct answer is

Option C) 5.62 points

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2 years ago
Factorise fully<br>-12a - 16 <br><br>please do quick, first to answer marked brainiest!!​
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Answer:

-4( 3a+4)

Step-by-step explanation:

-12a - 16

-4*3a -4*4

Factor out a -4

-4( 3a+4)

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3 years ago
A certain CD-ROM disk can store approximately 6.0 × 102 megabytes of information, where 106 bytes = 1 megabyte. If an average wo
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Answer: A. 4.3\times10^7

Step-by-step explanation:

Given : A certain CD-ROM disk can store approximately 6.0 \times 10^2  megabytes of information.

1 megabyte = 10^6 bytes

Then, The storage capacity of CD-ROM disk will be :

10^6\times 6.0 \times 10^2=6\times10^{6+2}=6\times10^8  

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\dfrac{6\times10^8}{14}=0.428571428571\times10^8\approx0.43\times10^8=4.3\times10^7

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4 0
3 years ago
A random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6. A random sample of 17 su
Sladkaya [172]

Answer:

We conclude that there is no difference in potential mean sales per market in Region 1 and 2.

Step-by-step explanation:

We are given that a random sample of 12 supermarkets from Region 1 had mean sales of 84 with a standard deviation of 6.6.

A random sample of 17 supermarkets from Region 2 had a mean sales of 78.3 with a standard deviation of 8.5.

Let \mu_1 = mean sales per market in Region 1.

\mu_2  = mean sales per market in Region 2.

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0      {means that there is no difference in potential mean sales per market in Region 1 and 2}

Alternate Hypothesis, H_A : > \mu_1-\mu_2\neq 0      {means that there is a difference in potential mean sales per market in Region 1 and 2}

The test statistics that will be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;

                            T.S.  =  \frac{(\bar X_1 -\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+ {\frac{1}{n_2}}} }   ~  t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean sales in Region 1 = 84

\bar X_2 = sample mean sales in Region 2 = 78.3

s_1  = sample standard deviation of sales in Region 1 = 6.6

s_2  = sample standard deviation of sales in Region 2 = 8.5

n_1 = sample of supermarkets from Region 1 = 12

n_2 = sample of supermarkets from Region 2 = 17

Also, s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times  s_2^{2}  }{n_1+n_2-2} }  = s_p=\sqrt{\frac{(12-1)\times 6.6^{2}+(17-1)\times  8.5^{2}  }{12+17-2} } = 7.782

So, <u><em>the test statistics</em></u> =  \frac{(84-78.3)-(0)}{7.782 \times \sqrt{\frac{1}{12}+ {\frac{1}{17}}} }  ~   t_2_7

                                   =  1.943  

The value of t-test statistics is 1.943.

 

Now, at a 0.02 level of significance, the t table  gives a critical value of -2.472 and 2.473 at 27 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have<u><em> insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that there is no difference in potential mean sales per market in Region 1 and 2.

6 0
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Answer:10x^2y^2

Step-by-step explanation:

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