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ira [324]
3 years ago
8

Find the sample size necessary in order to be 95% confident when determining the true mean weight within 2 units (EC2). Assume t

hat sample variance is 44. a. 65 b. 157 c. 43 d. 170 e. 89
Mathematics
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

(c) 43.

Step-by-step explanation:

Given:

Margin of error (E) = 2

Sample variance (s^{2}) = 44

Confidence level = 95%

The z- value at 95% confidence level is 1.96.

The sample using the provided information can be calculated as:

n=(\frac{z(s)}{E})^{2} \\n=\frac{z^{2} s^{2} }{E^{2} } \\n=\frac{1.96^{2} (44) }{2^{2} }\\n= 42.52\\n=43

Hence, the correct option is (c) 43.

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Which type of checking account requires a certain amount of money to be
lisabon 2012 [21]

Answer:

the answer is a

Step-by-step explanation:

minimum balance account

4 0
3 years ago
Read 2 more answers
Yoxel buys 4 1/2 gallons of soda. 1/4of the soda he bought was Pepsi and the rest was Sprite. How many gallons of Pepsi did Yoxe
marusya05 [52]

Answer:

4\frac{1}{4} gallons

Explanation:

Total amount of soda bought by Yoxel = 4\frac{1}{2} =\frac{9}{2} gallons

Amount of Pepsi =\frac{1}{4} gallons

Therfore,

Amount of sprite ==\frac{9}{2} -\frac{1}{4}\\=\frac{18-1}{4} \\=\frac{17}{4}\\ =4\frac{1}{4}gallons

8 0
3 years ago
345 in standard form
steposvetlana [31]
345 in Standard Form is 3.45 x 10^2

This is because when in standard form the decimal should move to the right 2 times.
5 0
3 years ago
A recent study from the University of Virginia looked at the effectiveness of an online sleep therapy program in treating insomn
Ludmilka [50]

Answer:

1. The 99% confidence interval for the difference in average is -6.47377 < μ₁ - μ₂ < -11.34623

2. The possible issues in the calculations includes;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

Step-by-step explanation:

1. The number of adults with insomnia in the sample = 45

The number of adults that participated in the therapy, n₁ = 22

The number of candidates that served as control group, n₂ = 23

The average score for the for the 22 participants of the program, \overline x_1 = 6.59

The standard deviation for the 22 participants of the program, s₁ = 4.10

The average score for the for the 23 subjects in the control group, \overline x_2 = 15.50

The standard deviation for the 23 subjects in the control group, s₂ = 5.34

The confidence interval for unknown standard deviation, σ, is given by the following expression;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

α = 1 - 0.99 = 0.01

α/2 = 0.005

The degrees of freedom, df = 22 - 1 = 21

t_{\alpha /2} = t_{0.005, \, 21} = 1.721

Therefore, we have;

\left (6.59- 15.5  \right )\pm1.721 \cdot \sqrt{\dfrac{4.10^{2}}{22}+\dfrac{5.34^{2}}{23}}

The 99% confidence interval for the difference in average is therefore given as follows;

-6.47377 < μ₁ - μ₂ < -11.34623

Therefore, there is considerable evidence that the participants in the survey  had lower average score than the subjects in the control group

2. The possible issues in the calculations are;

a. The confidence level used in the confidence interval can influence the result of the confidence interval observed

b. The sample size is small

5 0
2 years ago
Complete the pattern
Mademuasel [1]
38 x 1 = 1
38 x 0.1 = 3.8
38 x 0.01 = 0.38
8 0
4 years ago
Read 2 more answers
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