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2 years ago

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The electric cooperative needs to know the mean household usage of electricity by its non-commercial customers in kWh per day. T

hey would like the estimate to have a maximum error of 0.15 kWh. A previous study found that for an average family the standard deviation is 1.9 kWh and the mean is 16.7 kWh per day. If they are using a 98% level of confidence, how large of a sample is required to estimate the mean usage of electricity? Round your answer up to the next integer

1 answer:

Bogdan [553]2 years ago

7 0

Answer: 263

Step-by-step explanation:

As per given , we have

Population standard deviation : $\sigma=1.9\text{ kWh}$

maximum error : E=0.15 kWh

Significance level : $\alpha=1-0.98=0.02$

Critical value for 98% confidence interval : $z_{\alpha/2}=1.28$

Formula to find the sample size :

$n=(\dfrac{z_{\alpha/2}\cdot \sigma}{E})^2\\\\=(\dfrac{1.28\times1.9}{0.15})^2\\\\=262.872177778\approx263$

Hence, the minimum sample size required =263

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A study was recently conducted at a major university to estimate the difference in the proportion of business school graduates w

sveta [45]

Answer:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p_A$ represent the real population proportion for business

$\hat p_A =\frac{75}{400}=0.1875$ represent the estimated proportion for Business

$n_A=400$ is the sample size required for Business

$p_B$ represent the real population proportion for non Business

$\hat p_B =\frac{137}{500}=0.274$ represent the estimated proportion for non Business

$n_B=500$ is the sample size required for non Business

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.1875-0.274) - 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.1412$

$(0.1875-0.274) + 1.96 \sqrt{\frac{0.1875(1-0.1875)}{400} +\frac{0.274(1-0.274)}{500}}=-0.0318$

And the 95% confidence interval would be given (-0.1412;-0.0318).

We are confident at 95% that the difference between the two proportions is between $-0.1412 \leq p_A -p_B \leq -0.0318$

7 0

1 year ago

A turtle swims at a speed of 1584 meters per hour. A girl swims at a speed of 1380 millimeters per second how much faster does t

d1i1m1o1n [39]

The turtle swims faster. In fact, the turtle swims 812 cm/min faster.

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1 year ago

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The keyless entry system in some cars uses a 4-digit keypad. There are 10 possible digits, and the digits can be repeated. How m

Len [333]

This item can be answered through the concept of fundamental principles of counting. In the first of the four digits, there are 10 possible digits. The same with all the other 3 places or digits of the code. That is,
n = 10 x 10 x 10 x 10
Giving us the answer of 10,000. Thus, there are 10,000 possible keys for the keyless entry.

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2 years ago

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The length of time a full length movie runs from opening to credits is normally distributed with a mean of 1.9 hours and standar

Llana [10]

Answer:

a) The probability that a random movie is between 1.8 and 2.0 hours = 0.2586.

b) The probability that a random movie is longer than 2.3 hours is 0.0918.

c) The length of movie that is shorter than 94% of the movies is 1.4 hours

Step-by-step explanation:

In the above question, we would solve it using z score formula

z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation

a) A random movie is between 1.8 and 2.0 hours

z = (x-μ)/σ,

x1 = 1.8,

x2 = 2.0

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z1 = (1.8 - 1.9)/0.3

z1 = -1/0.3

z1 = -0.33333

Using the z score table

P(z1 = -0.33) = 0.3707

z2 = (2.0 - 1.9)/0.3

z1 = 1/0.3

z1 = 0.33333

p(z2 = 0.33) = 0.6293

= P(- 0.33 ≤ z ≤ 0.33)

= 0.6293 - 0.3707

= 0.2586

The probability that a random movie is between 1.8 and 2.0 hours = 0.2586

b) A movie is longer than 2.3 hours

z = (x-μ)/σ,

x1 = 2.3

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

z = (2.3 - 1.9)/0.3

z = 4/0.3

z = 1.33333

P(z = 1.33) = 0.90824

P(x>2.3) = = 1 - 0.90824

= 0.091759

≈ 0.0918

The probability that a random movie is longer than 2.3 hours is 0.0918.

3) The length of movie that is shorter than 94% of the movies.

z = (x-μ)/σ

Probability (z ) = 94% = 0.94

Movie that is shorter than 0.94

= P(1 - 0.94) = P(0.06)

Finding the P (x< 0.06) = -1.555

≈ -1.56

μ is the population mean = 1.9

σ is the population standard deviation = 0.3

-1.56 = (x - 1.9)/ 0.3

Cross multiply

-1.56 × 0.3 = x - 1.9

- 0.468 + 1.9 = x

= 1.432 hours

≈ 1.4 hours

Therefore, the length of movie that is shorter than 94% of the movies is 1.4 hours

5 0

1 year ago

The inequality that will determine the number of months, x, that are required for the second phone to be less expensive is . The

ziro4ka [17]

<em>Question:</em>

<em>Sal is trying to determine which cell phone and service plan to buy for his mother. The first phone costs $100 and $55 per month for unlimited usage. The second phone costs $150 and $51 per month for unlimited usage. </em>

<em>The inequality that will determine the number of months, x, that are required for the second phone to be less expensive is . </em>

<em>The solution to the inequality is . </em>

<em>Sal’s mother would have to keep the second cell phone plan for at least months in order for it to be less expensive.</em>

Answer:

a. $150 + 51x < 100 + 55x$

b. $x > 12.5$

c. At least 13 months

Step-by-step explanation:

Given

First Phone;

$Cost = \$100$

$Additional = \$55$ <em>(monthly)</em>

Second Phone;

$Cost = \$150$

<em /> $Additional = \$51$ <em> (monthly)</em>

<em></em>

Solving (a): The inequality

<em></em>

Represent the number of months with x

The first phone is expressed as:

$100 + 55x$

The second phone is expressed as:

$150 + 51x$

For the second to be less expensive that the first, the inequality is:

$150 + 51x < 100 + 55x$

Solving (b): Inequality Solution

$150 + 51x < 100 + 55x$

Collect Like Terms

$51x-55x$

$-4x$

Solve for x

$x > -50/-4$

$x > 12.5$

Solving (c): Interpret the solution in (b)

$x > 12.5$ implies 13, 14, 15....

Hence, She'll keep the second phone for a period of at least 13 months

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1 year ago

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