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-BARSIC- [3]
2 years ago
11

g Suppose the population 10-year cumulative incidence of prostate cancer among 70-year old men is 0.06. Further, suppose 260 men

without prostate cancer, aged 50 years were randomly sampled and 13 of them developed prostate cancer within 10 years. Is there evidence at the 0.05 significance level to conclude that the cumulative incidence of prostate cancer among 50-year old men differs from that of 70-year old men
Mathematics
1 answer:
Rainbow [258]2 years ago
5 0

Answer:

There is not enough statistical evidence to conclude that the cumulative incidence of prostrate cancer among 50-year old men differs from that of 70-year old men

Step-by-step explanation:

The population 10 year cumulative incidence of prostrate cancer among 70 year old men, p = 0.06

The number of 50 year old men in the sample, n = 260 men

The number of the sampled men that developed prostrate cancer = 13 men

The significance level, α = 0.05

Let the null hypothesis, H₀: \hat{p} = p

The alternative hypothesis, Hₐ: \hat{p} ≠ p

The standard score, z_{\alpha /2} = 1.96

The test statistic, 'z', is given as follows;

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p \cdot q}{n}}}

\hat{p} = 13/260 = 0.05

q = 1 - p = 1 - 0.06 = 0.94

Therefore,  \ z=\dfrac{0.05-0.06}{\sqrt{\dfrac{0.06 \times 0.94}{260}}} = \dfrac{\sqrt{9165} }{141} \approx -0.68

From the z-table, we find the p-value as follows;

P(z ≈ -0.68) = 0.24825

Therefore, given that the p-value, 0.24825 is larger than the significance level, α/2 = 0.025, we fail to reject the null hypothesis, and therefore, there is not enough statistical evidence to conclude that the cumulative incidence of prostrate cancer among 50-year old men differs from that of 70-year old men

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To find these percents, you need to find the z-score for each value. Then, use your table to find the correct percent. Be sure to find the side above 13 when you use your chart.

For less than 3:
(3 - 7.45) / 3.6 = -1.24 = The percent below this is 0.1075

For greater than 13:
(13 - 7.45) / 3.6 = 1.54 = The percent above this is 0.0618
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Can you answer these for me ?
vodka [1.7K]

Answer:

B. 4 3/8

Step-by-step explanation:

7/8 X 5 = 4 3/8

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3 years ago
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For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger
11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let <em>X</em> = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = p=\frac{1}{120}.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X </em>is:

P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...

In this case we need to compute the probability of 1 or more than 1 catastrophes in <em>n</em> missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             =1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419

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stepladder [879]

Answer:

The answer is d) 480cm³

Step-by-step explanation:

3 x 10 x 16 = 480cm³

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