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lord [1]
3 years ago
15

An investigator wants to estimate caffeine consumption in high school students. how many students would be required to estimate

the proportion of students who consume coffee? suppose we want the estimate to be within 5% of the true proportion with 95% confidence.
Mathematics
1 answer:
astraxan [27]3 years ago
6 0

We can solve this problem by referring to the standard probability distribution tables for z.

We are required to find for the number of samples given the proportion (P = 5% = 0.05) and confidence level of 95%. This would give a value of z equivalent to:

z = 1.96

Since the problem states that it should be within the true proportion then p = 0.5

Now we can find for the sample size using the formula:

n = (z^2) p q /E^2

where,

<span> p = 0.5</span>

q = 1 – p = 0.5

E = estimate of 5% = 0.05

Substituting:

n = (1.96^2) 0.5 * 0.5 / 0.05^2

n = 384.16

<span>Around 385students are required.</span>

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The number of typing errors made by a typist has a Poisson distribution with an average of seven errors per page. If more than s
Aleks [24]

Answer:

0.599  is the probability that a randomly selected page does not need to be retyped.

Step-by-step explanation:

We are given the following in the question:

The number of typing errors made by a typist has a Poisson distribution.

\lambda  =7

The probability is given by:

P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}

We have to find the probability that a  randomly selected page does not need to be retyped

P(less than or equal to 7 mistakes in a page)

P( x \leq 7) = P(x =1) + P(x=1) +...+ P(x=6) + P(x = 7)\\\\= \displaystyle\frac{7^0 e^{-7}}{0!} + \displaystyle\frac{7^1 e^{-7}}{1!} +...+ \displaystyle\frac{7^6 e^{-7}}{6!} + \displaystyle\frac{7^7 e^{-7}}{7!} \\\\= 0.59871\approx 0.599

Thus, 0.599  is the probability that a randomly selected page does not need to be retyped.

4 0
3 years ago
In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
uysha [10]

Answer:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

Step-by-step explanation:

For this problem we have the following data given:

\bar X_1 = 9 represent the sample mean for one of the departments

\bar X_2 = 8 represent the sample mean for the other department

n_1 = 25 represent the sample size for the first group

n_2 = 20 represent the sample size for the second group

s_1 = 2 represent the deviation for the first group

s_2 =1 represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

The confidence given is 95% or 9.5, then the significance level is \alpha=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n_1 +n_2 -2= 20+25-2= 43

And the critical value for this case is t_{\alpha/2}=2.02

And replacing we got:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

4 0
3 years ago
A temperature recorded in Antarctica was −123°F. The temperature recorded in the Sahara Desert was 135°F. How many degrees warme
Mashcka [7]

Answer:

<em>The Sahara Desert is 258 °F warmer that Antarctica.</em>

<em></em>

Step-by-step explanation:

Temperature at Antarctica = -123 °F

Temperature at the Sahara desert = 135 °F

The difference in temperature = 135 - (-123) = 135 + 123 = 258 °F

This means that the Sahara Desert is 258 °F warmer that Antarctica.

8 0
4 years ago
If g(x) = (7x3 - 1)14(x2 + 50)49, what is g '(x)? a [14(7x3 - 1)13][49(x2 + 50)48] b14(7x3 - 1)13(21x2) + 49(x2 + 50)48(2x) c 14
Yanka [14]
<span>b.14(7x3 - 1)13(21x2) + 49(x2 + 50)48(2x)</span>
5 0
3 years ago
Jack estimates that the cost per mile, in dollars, for operating a certain truck is between 15% and 21% of the number of miles d
Xelga [282]

Answer:

  D)  If Jack drives 200 miles, it will cost anywhere between $30 and $42

Step-by-step explanation:

The cost is said to be a range of possibilities. The first three answer choices seem to assume the cost is at one extreme or the other. They incorrectly interpret the statement of cost.

7 0
3 years ago
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