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Alla [95]
2 years ago
15

Which of the following is NOT true when testing a claim about a​ proportion? Choose the correct answer below. A. Both the tradit

ional method and​ P-value method use the same standard deviation based on the claimed proportion​ p, but the confidence interval uses an estimated standard deviation based on the sample proportion ModifyingAbove p with caretp. B. When testing claims about population​ proportions, the traditional method and the​ P-value method are equivalent in the sense that they always yield the same result. C. A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test. D. If you want to test a claim about population​ proportions, use the​ P-value method or the classical method of hypothesis testing.
Mathematics
1 answer:
morpeh [17]2 years ago
5 0

Answer: C. A conclusion based on a confidence interval estimate will be the same as a conclusion based on a hypothesis test.

Explanation: The One-Sample Proportion Test is used to assess whether a population proportion (P1) is significantly different from a hypothesized value (P0). This procedure calculates sample size and statistical power for testing a single proportion using either the exact test or other approximate z-tests.

To write a null hypothesis, first, start by asking a question. Rephrase that question in a form that assumes no relationship between the variables. In other words, assume a treatment has no effect. Write your hypothesis in a way that reflects this.

A null hypothesis is a hypothesis that says there is no statistical significance between the two variables. It is usually the hypothesis a researcher or experimenter will try to disprove or discredit. An alternative hypothesis is one that states there is a statistically significant relationship between two variables.

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Triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 t
Ghella [55]

Answer:  The correct option is (D) P″(9, -12) and Q″(15, -3).

Step-by-step explanation:  Given that triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 to form triangle P″Q″R″.

The co-ordinates of vertices P and Q are (3, -4) and (5, -1) respectively.

We are to find the co-ordinates of the vertices P″ and Q″.

<u>Case I :</u>  ΔPQR dilated to ΔP'Q'R'

The co-ordinates of P' and Q' are given by

P'(3\times 1.5, -4\times 1.5)=P'(4.5, -6),\\\\Q'(5\times 1.5, -1\times 1.5)=Q'(7.5, -1.5).

<u>Case II :</u>  ΔP'Q'R' dilated to ΔP''Q''R''

The co-ordinates of P'' and Q'' are given by

P''(4.5\times 2, -6\times 2)=P'(9, -12),\\\\Q'(7.5\times 2, -1.5\times 2)=Q'(15, -3).

Thus, the co-ordinates of the vertices P'' and Q'' are (9, -12) and (15, -3).

Option (D) is CORRECT.

<u></u>

4 0
2 years ago
Read 2 more answers
Isosceles △ABC (AC=BC) is inscribed in the circle k(O). Prove that the tangent to the circle at point C is parallel to AB .
timofeeve [1]

Explanation:

Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.

If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.

5 0
2 years ago
From 2000-2003, students were tested by the state in four major subject areas of math, science, English and social studies. The
Verizon [17]

No way to tell . . . . . we can't see the chart below.
It must be WAY down there where the sun don't shine.
8 0
1 year ago
Se necesitan 4/7 de litro de pintura para pintar un metro cuadrado de pared, si queremos pintar 2/5 de metro cuadrado de pared,
Assoli18 [71]

Answer:

8/35 de litro

Step-by-step explanation:

De la pregunta, sabemos que se necesitan 4/7 de un litro para pintar un metro cuadrado de pared.

Ahora necesitamos saber la cantidad de pintura necesaria para pintar 2/5 de un metro cuadrado.

Digamos que la cantidad necesaria es x

Ahora hagamos una relación matemática;

4/7 pinturas 1 metro cuadrado

x pintará 2/5 metros

cuadrados por lo tanto, x * 1 = 4/7 * 2/5 = 8/35

6 0
2 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
2 years ago
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