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babymother [125]
3 years ago
13

A medical researcher says that less than 87​% of adults in a certain country think that healthy children should be required to

be vaccinated. In a random sample of 500 adults in that​ country, 85​% think that healthy children should be required to be vaccinated. At alphaequals0.05​, is there enough evidence to support the​ researcher's claim? Complete parts​ (a) through​ (e) below.
Mathematics
1 answer:
Lyrx [107]3 years ago
3 0

Answer: It is not enough evidence to support the researcher's claim.

Step-by-step explanation:

Since we have given that

n = 500

\hat{p}=0.85

So, the hypothesis are as follows:

H_0:\hat{p}

So, the z value would be

z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.85-0.87}{\sqrt{\dfrac{0.87\times 0.13}{500}}}\\\\z=\dfrac{-0.02}{0.015}\\\\z=-1.33

At α = 0.05

z= -1.64

So, 1.64<-1.33

So, we accept the null hypothesis.

Hence, it is not enough evidence to support the researcher's claim.

You might be interested in
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

Z = \frac{X - \mu}{s}

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

7 0
3 years ago
Spencer throws a fair six-sided dice.
sergejj [24]

Answer:

Step-by-step explanation:

P(E)=3/6=1/2

So it is equally probable to roll even as it is to roll an odd.

’evens’

5 0
3 years ago
(25 points)
kiruha [24]
The answer is A.

The 0's of the function are -2, 2, and 3.

The graph starts from the bottom left and it keeps going to the top right.

6 0
3 years ago
Read 2 more answers
-1 3/5 divided by -2/3 <br> Write as a mixed number
mihalych1998 [28]

Answer:

2 2/5

Step-by-step explanation:

Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction. First convert the first into the improper fraction -8/5. Now because when dividing fractions you keep the first on, change to multiplication and flip the second equation you have

-8/5 x -3/2=?

Now you just multiply across as you would and normal time multiplying fractions. Remember a negative divided by a negative is a positive! And not after you keep, change, flip and multiply you are left with 24/10. Which is equal to 12/5 but you wanted a mixed number so we convert it into 2 2/5

8 0
3 years ago
So how yall doing so whats 200 times 3/4 =
horrorfan [7]

Answer:

200 times 3/4 is 150

Hope this help!

Brainliest??

6 0
3 years ago
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