Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2)

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

Rainbow [258]

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

5 0

2 years ago

How do you solve this?

motikmotik

Pemdas helps so first you do Parentheses Exponet multiply divide add and subtract

3 0

2 years ago

Read 2 more answers

A maker of a certain brand of low-fat cereal barsclaims that the average saturated fat content is 0.5gram. In a random sample of

finlep [7]

Answer:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

$p_v =2*P(t_{(7)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

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".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=0.475$

The sample deviation calculated $s=0.183$

We need to conduct a hypothesis in order to check if the true mean is 0.5 or no, the system of hypothesis would be:

Null hypothesis: $\mu = 0.5$

Alternative hypothesis: $\mu \neq 0.5$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{0.475-0.5}{\frac{0.183}{\sqrt{8}}}=-0.386$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=8-1=7$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(7)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is different from 0.5 at 5% of signficance. So then the claim makes sense

4 0

3 years ago

Read 2 more answers

LaDanian wants to write the fraction 4 / 6 as a sum of unit fractions. what expression should he write

julsineya [31]

He <span>wants to write the fraction 4 / 6 as a sum of unit fractions. It is written as follows:
</span>
1/6 +1/6 +1/6+ 1/6= 4/6.

A unit fraction has a numerator of 1 so since 4 is the numerator in 4/6 you would have to add 1/6 for four times.

Hope this answers the question. Have a nice day.

7 0

3 years ago

Hey, I need help with questions 1 and 2, if anyone can help, please? thanks!

Eva8 [605]

The correct works are: