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2 years ago

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Natalie surveyed 800 of the students in her school about their favorite color. Students could choose between blue and red. 77% s

aid their favorite color was red. How many students' favorite color was blue?

2 answers:

melamori03 [73]2 years ago

6 0

184 students said their favorite color was blue. 77% of 800 is 616, then you subtract 800-616 and get 184.

matrenka [14]2 years ago

3 0

To solve this question you would do 800 Times 77%, because 77% of the 800 children favorite color was red. So that mean 616 students favorite color was red and 800-616= 184. So 184 students favorite color was blue

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A random sample of 38 wheel chair users were asked whether they preferred cushion type A or B, and 28 of them preferred type A w

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The statement that cushion A is twice as popular as cushion B cannot be verified

Step-by-step explanation:

From the question we are told that:

Sample size n=38

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Type a size B $X_b=10$

Generally the probability of choosing cushion A P(a) is mathematically given by

$P(a)=\frac{28}{38}$

$P(a)=0.73$

Generally the equation for A to be twice as popular as B is mathematically given by

$P(b)+2P(b)=3P$

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$Null H_0: p \leq \frac{2P}{3P} \\Altenative H_A:p>\frac{2P}{3P}$

Generally the equation normal approx of p value is mathematically given by

$z=\frac{x-np_0-0.5}{\sqrt{np_0(1-p_0)} }$

$z=\frac{28-(38*2/3)_0-0.5}{\sqrt{38*2/3*1/3} }$

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Therefore from distribution table

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Therefore the statement that cushion A is twice as popular as cushion B cannot be verified

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$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

being the subindex 1 for the experimental group and subindex 2 for the control group.

2) Test statistic z=-1.26

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the vaccine was effective.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2< 0$

The significance level is 0.01.

The sample 1 (experimental group), of size n1=100000 has a proportion of p1=0.0002.

$p_1=X_1/n_1=21/100000=0.0002$

The sample 2 (control group), of size n2=100000 has a proportion of p2=0.0003.

$p_2=X_2/n_2=30/100000=0.0003$

The difference between proportions is (p1-p2)=-0.0001.

$p_d=p_1-p_2=0.0002-0.0003=-0.0001$

The pooled proportion, needed to calculate the standard error, is:

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$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.00026*0.99974}{100000}+\dfrac{0.00026*0.99974}{100000}}\\\\\\s_{p1-p2}=\sqrt{0+0.0000000025}=\sqrt{0.0000000051}=0.000071$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.00009-0}{0.000071}=\dfrac{-0.00009}{0.000071}=-1.26$

This test is a left-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=P(z$

As the P-value (0.104) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the vaccine was effective.

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