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3 years ago

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Testing to see if there is evidence that the proportion of US citizens who can name the capital city of Canada is greater than 0

.75. a. (3pts) State the Null and Alternative Hypotheses. Also, state the direction of the test Using the following possible sample results: • Sample A: 31 successes out of 40 • Sample B: 34 successes out of 40 • Sample C: 27 successes out of 40 • Sample D: 38 successes out of 40 b. State which of the possible sample results provides the most evidence against the null. Briefly explain (2pts) c. State which (if any) of the possible results provide no evidence for the alternative. Briefly explain (2pts)

1 answer:

klemol [59]3 years ago

8 0

Answer:

b) B and D and c) A and C

Step-by-step explanation:

The hypotheses would be:

$H_0: p =0.75\\H_a: p >0.75$

(Right tailed test at 5% level)

Sample proportions are

Sample A B C D

Success 31 34 27 38

Proportion p 0.775 0.85 0.675 0.95

Std error

(sqrtpq/n) 0.066 0.056 0.074 0.034

p diff

p-0.75 0.025 0.10 - 0.075 0.20

Z stat

p diff/SE 0.0757 1.77 1.01 5.80

p value 0.469 0.038 0.156 0.000001

we find that p value is more smaller than alpha, the more accurate the alternate hypothesis.

b) Only B and D provide against the null. Because p <0.05

c) A and C provide no evidence for the alternative because p >0.05

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Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the

max2010maxim [7]

Answer:

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so $n = 972, \pi = \frac{272}{972} = 0.28$

85% confidence level

So $\alpha = 0.15$ , z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$ , so $Z = 1.44$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301$

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

3 0

3 years ago

Can you answer this one for my plz I need before Monday ty

ira [324]

Since you need the vertex, and knowing the vertex will tell you the axis of symmetry, it is convenient to put the equation in vertex form.
.. -x^2 -4x +1
.. = -(x^2 +4x) +1
.. = -(x^2 +4x +(4/2)^2) +1 -(-(4/2)^2) . . . . complete the square
.. = -(x +2)^2 +5

The vertex is (-2, 5).
The axis of symmetry is x = -2.

7 0

3 years ago

(4.61×10^4)-(2.75×10^4)

Yanka [14]

(46100)-(27500)=18600

6 0

3 years ago

What is the equation of the line that passes through (0, 3) and (7, 0)?

r-ruslan [8.4K]

The answer should be (C):y=(-3/7)x+3 because:

if you use the slope formula: (Y2-Y1)/(X2-X1) it would look like this:

(0-3)/(7-0) and the slope of the line would be -3/7

then using the slope you just found, plug it into the y=mx+b equation along with one set of coordinates to find b (the y-intercept):

y=mx+b
0=(-3/7)(7)+b
b=3

SO, overall if you put the slope (m) and the y-intercept (b) that you found together, you get the line formula of y=(-3/7)+3

5 0

3 years ago

Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lara31 [8.8K]

Answer:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

$X=35$ number of people that agree.

$n=100$ random sample taken

$\hat p=\frac{35}{100}=0.35$ estimated proportion of people that agree

$p$ true population proportion of peopl that agree

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".

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The standard error is given by:

$SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048$

If we replace the values obtained we got:

$0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257$

$0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443$

The 95% confidence interval would be given by (0.257;0.443)

6 0

4 years ago

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