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solniwko [45]
3 years ago
5

Consider random samples of size 58 drawn from population A with proportion 0.77 and random samples of size 70 drawn from populat

ion B with proportion 0.67 . (a) Find the standard error of the distribution of differences in sample proportions, . Round your answer for the standard error to three decimal places. standard error = Enter your answer in accordance to the question statement (b) Are the sample sizes large enough for the Central Limit Theorem to apply?
Mathematics
1 answer:
fredd [130]3 years ago
8 0

Answer:

Step-by-step explanation:

a) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as

Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}

where

p1 = sample proportion of population 1

p2 = sample proportion of population 2

n1 = number of samples in population 1,

n2 = number of samples in population 2,

From the information given

p1 = 0.77

1 - p1 = 1 - 0.77 = 0.23

n1 = 58

p2 = 0.67

1 - p2 = 1 - 0.67 = 0.33

n2 = 70

Standard error = √{(0.77 - 0.67)/[(0.77)(0.23)/58) + (0.67)(0.33)/70}

= √0.1/(0.0031 + 0.0032)

= √1/0.0063

= 12.6

the standard error of the distribution of differences in sample proportions is 12.6

b) the sample sizes are large enough for the Central Limit Theorem to apply because it is greater than 30

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Answer:

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In AOPQ, o = 700 cm, p = 840 cm and q=620 cm. Find the measure of _P to the<br> nearest degree
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Given:

In triangle OPQ, o = 700 cm, p = 840 cm and q=620 cm.

To find:

The measure of angle P.

Solution:

According to the Law of Cosines:

\cos A=\dfrac{b^2+c^2-a^2}{2bc}

Using Law of Cosines in triangle OPQ, we get

\cos P=\dfrac{o^2+q^2-p^2}{2oq}

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On further simplification, we get

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P=78.786236

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Therefore, the measure of angle P is 79 degrees.

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3 years ago
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Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation o
hjlf

Answer:

46.18% of the items will weigh between 6.4 and 8.9 ounces.

Step-by-step explanation:

We are given that the weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.

<em>Let X =  weight of items produced by a machine</em>

The z-score probability distribution for is given by;

                Z = \frac{  X -\mu}{\sigma}  ~ N(0,1)

where, \mu = mean weight = 8 ounces

            \sigma = standard deviation = 2 ounces

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, percentage of items that will weigh between 6.4 and 8.9 ounces is given by = P(6.4 < X < 8.9) = P(X < 8.9 ounces) - P(X \leq 6.4 ounces)

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