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ivann1987 [24]
2 years ago
12

A survey of 132 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take

notes. The result of the survey is that 66 of the 132 students responded "yes.". An approximate 98% confidence interval is (0.399, 0.601). How would the confidence interval change if the confidence level had been 90% instead of 98%
Mathematics
1 answer:
ddd [48]2 years ago
7 0

Answer:

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

Given that;

Proportion p = 66/132 = 0.50

Number of samples n = 132

Confidence level = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

0.50 +/- 1.645√(0.50(1-0.50)/132)

0.50 +/- 1.645√(0.001893939393)

0.50 +/- 0.071589436011

0.50 +/- 0.072

(0.428, 0.572)

The 90% confidence level estimate of the true population proportion of students who responded "yes" is (0.428, 0.572)

For 90% CI = (0.428, 0.572)

For 98% CI = (0.399, 0.601)

The confidence interval (and Margin of error) reduces when 90% confidence level is used compared to when 98% confidence level is used.

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This side of square S is twice the diameter of circle C what is the greatest possible number of points of intersection of perime
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Answer:

2

Step-by-step explanation:

C is too small to have more.

5 0
1 year ago
18.) The cost of textbooks for a school increases with the average class size. Identify the independent and dependent quantity i
son4ous [18]

The independent quantity is the average class size and the dependent quantity is the cost ⇒ c

Step-by-step explanation:

Lets revise the meaning of dependent and independent variables

  • The dependent variable is the one that depends on the value of some other number
  • The dependent variable is the output value and the independent variable is the input value
  • Ex: if y = 2x + 3, x is the input and y is the out put, then is independent variable and y is dependent variable

∵ The cost of the textbook increases with the average class size

- The cost of the textbooks depends on the average of the

  class size

∴ The input is the average class size

∴ The output is the cost of the textbooks

∵ The input variable is independent variable

∵ The output variable is dependent variable

∴ The average of the class size is independent

∴ The cost of the textbooks is dependent

The independent quantity is the average class size and the dependent quantity is the cost

Learn more:

You can learn more about the word problems in brainly.com/question/13174281

#LearnwithBrainly

8 0
2 years ago
A warehouse has boxes that are 18 inches tall and other boxes that are 30 inches tall. If 18 inch boxes is stacked next to 30 in
Aleksandr-060686 [28]

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4 0
2 years ago
You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the
Makovka662 [10]

Answer:

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

  • The total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

Step-by-step explanation:

a)  How much will you have at the middle of the first year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

To determine:

Total amount = A = ?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

substituting the values

A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}

A=300\cdot \frac{2.06}{2}

A=\frac{618}{2}

A=309 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

where

  • Principle = P
  • Annual rate = r
  • Compound = n
  • Time  = (t in years)
  • A = Total amount

Given:

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

To determine:

Total amount = A = ?

so using the formula

A\:=\:P\left(1+\frac{r}{n}\right)^{nt}

so substituting the values

A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}

A=300\cdot \frac{2.06^2}{2^2}

A=318.27 $

Therefore, the total amount accrued, principal plus interest,  from compound interest on an original principal of  $ 300.00 at a rate of 6% per year  compounded 2 times per year  over 1 year is $ 318.27.

3 0
2 years ago
Suppose that 90% of all dialysis patients will survive for at least 5 years. In a simple random sample of 100 new dialysis patie
Keith_Richards [23]

Answer:

The  probability   P(\^ p  >  0.80) =  0.99957

Step-by-step explanation:

From the question we are told that

     The population proportion is  p  =  0.90

     The sample size is  n  =  30

  Generally mean of the sampling distribution is  \mu_{\= x } =  p =  0.90

Generally the standard deviation is mathematically represented as

      \sigma  = \sqrt{\frac{p( -p )}{n} }

=>    \sigma  = \sqrt{\frac{0.90( 1 -0.90 )}{100} }

=>    \sigma  = 0.03

Generally the he probability that the proportion surviving for at least five years will exceed 80%, rounded to 5 decimal places is mathematically represented as

       P(\^ p  >  0.80) =  P(\frac{ \^ p - p }{ \sigma }  >  \frac{0.80 - 0.90}{0.03} )

Generally \frac{\^p - p}{\sigma}  =  Z(The standardized \  value  \  of  \^  p)

So  

    P(\^ p  >  0.80) =  P(Z  >  -3.33 )

From the z-table P(Z  >  -3.33 ) = 0.99957

So

    P(\^ p  >  0.80) =  0.99957

5 0
2 years ago
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