Solve the differential equation. y ' = 9x

Suppose that a particular candidate for public office is in fact favored by 48%of all registered voters in the district. A polli

Mumz [18]

Answer: 0.1854

Step-by-step explanation:

Given : Suppose that a particular candidate for public office is in fact favored by 48% of all registered voters in the district.

Let $\hat{p}$ be the sample proportion of voters in the district favored a particular candidate for public office .

A polling organization will take a random sample of n=500 voters .

Then, the probability that p will be greater than 0.5, causing the polling organization to incorrectly predict the result of the upcoming election :

$P(\hat{p}>0.5)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.5-0.48}{\sqrt{\dfrac{0.48(0.52)}{500}}})\\\\=P(z>0.8951)\ \ [\because\ z=(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}]\\\\=1-P(z\leq0.8951)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\ = 1-0.8146=0.1854$

∴ Required probability = 0.1854

7 0

3 years ago

A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selecti

Andrews [41]

Answer:

(a) Null Hypothesis, $H_0$ : p = 0.50

Alternate Hypothesis, $H_A$ : p > 0.50

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed.

(e) The value of z test statistics is 0.96.

(f) The P-value is 0.1685.

(g) At 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) We conclude that the proportion of baby girls is equal to 0.50.

Step-by-step explanation:

We are given that a 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5.

Assume that sample data consists of 78 girls in 144 births.

Let p = population proportion of baby girls

(a) So, Null Hypothesis, $H_0$ : p = 0.50 {means that the proportion of baby girls is equal to 0.50}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the proportion of baby girls is greater than 0.50}

(b) The value of level of significance (α) given in the question is 0.10.

(c) The sampling distribution of the sample statistic is Normal distribution.

(d) This test is right-tailed as in the alternative hypothesis we are concerned for proportion of baby girls that is greater than 0.50.

(e) The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of baby girls = $\frac{78}{144}$ = 0.54

n = sample of births = 144

So, the test statistics = $\frac{0.54-0.50}{\sqrt{\frac{0.54(1-0.54)}{144} } }$

= 0.96

The value of z test statistics is 0.96.

(f) The P-value of the test statistics is given by;

P-value = P(Z > 0.96) = 1 - P(Z < 0.96)

= 1 - 0.8315 = 0.1685

(g) Now, at 0.10 significance level the z table gives critical value of 1.282 for right-tailed test.

(h) Since our test statistic is less than the critical value of z as 0.96 < 1.282, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region (which was to the right of value of 1.282) due to which we fail to reject our null hypothesis.

Therefore, we conclude that the proportion of baby girls is equal to 0.50.

7 0

3 years ago

Which of the following is the square of a binomial?

Advocard [28]

ANSWER

D.

$16 {x}^{2} + 24xy + 9 {y}^{2}$