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Tanzania [10]
3 years ago
5

A paper reported that in a representative sample of 282 American teens age 16 to 17, there were 71 who indicated that they had s

ent a text message while driving. For purposes of this exercise, assume that this sample is a random sample of 16- to 17-year-old Americans. Do these data provide convincing evidence that more than a quarter of Americans age 16 to 17 have sent a text message while driving? Test the appropriate hypotheses using a significance level of 0.01. (Round your test statistic to two decimal places and your P-value to four decimal places.)
Mathematics
1 answer:
elena-s [515]3 years ago
8 0

Answer:

z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776  

p_v =P(z>0.0776)=0.4691  

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

Step-by-step explanation:

Information provided

n=282 represent the sample selected

X=71 represent the teens indicated that they had sent a text message while driving

\hat p=\frac{71}{282}=0.252 estimated proportion of teens indicated that they had sent a text message while driving

p_o=0.25 is the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

System of hypothesis

We want to check if more than a quarter of Americans age 16 to 17 have sent a text message while driving, and the hypothesis are:

Null hypothesis:p\leq 0.7  

Alternative hypothesis:p > 0.25  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing we got:

z=\frac{0.252 -0.25}{\sqrt{\frac{0.25(1-0.25)}{282}}}=0.0776  

Decision

The p value for this case is:

p_v =P(z>0.0776)=0.4691  

Since the p value is higher than the significance level so then we can conclude that we have enough evidence to FAIL to reject the null hypothesis and there is no evidence in order to say that the true proportion of teens more than a quarter of Americans age 16 to 17 have sent a text message while driving is higher than 0.25

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If there is a song that is 2 minutes and 58 seconds long and it plays 40 times how long does it play convert answer into seconds
lapo4ka [179]

Answer: I got 7120 seconds

Step-by-step explanation:

There's 178 seconds in the song (for more context, I just converted the 2 minutes into seconds and added it with 58) then I multiplied it by the total amount of times it played (in this case, 40)

Hope this helps

3 0
3 years ago
Read 2 more answers
The mean number of words per minute (WPM) read by sixth graders is 8888 with a standard deviation of 1414 WPM. If 137137 sixth g
Bingel [31]

Noticing that there is a pattern of repetition in the question (the numbers are repeated twice), we are assuming that the mean number of words per minute is 88, the standard deviation is of 14 WPM, as well as the number of sixth graders is 137, and that there is a need to estimate the probability that the sample mean would be greater than 89.87.

Answer:

"The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

Step-by-step explanation:

This is a problem of the <em>distribution of sample means</em>. Roughly speaking, we have the probability distribution of samples obtained from the same population. Each sample mean is an estimation of the population mean, and we know that this distribution behaves <em>normally</em> for samples sizes equal or greater than 30 \\ n \geq 30. Mathematically

\\ \overline{X} \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [1]

In words, the latter distribution has a mean that equals the population mean, and a standard deviation that also equals the population standard deviation divided by the square root of the sample size.

Moreover, we know that the variable Z follows a <em>normal standard distribution</em>, i.e., a normal distribution that has a population mean \\ \mu = 0 and a population standard deviation \\ \sigma = 1.

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} [2]

From the question, we know that

  • The population mean is \\ \mu = 88 WPM
  • The population standard deviation is \\ \sigma = 14 WPM

We also know the size of the sample for this case: \\ n = 137 sixth graders.

We need to estimate the probability that a sample mean being greater than \\ \overline{X} = 89.87 WPM in the <em>distribution of sample means</em>. We can use the formula [2] to find this question.

The probability that the sample mean would be greater than 89.87 WPM

\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{89.87 - 88}{\frac{14}{\sqrt{137}}}

\\ Z = \frac{1.87}{\frac{14}{\sqrt{137}}}

\\ Z = 1.5634 \approx 1.56

This is a <em>standardized value </em> and it tells us that the sample with mean 89.87 is 1.56<em> standard deviations</em> <em>above</em> the mean of the sampling distribution.

We can consult the probability of P(z<1.56) in any <em>cumulative</em> <em>standard normal table</em> available in Statistics books or on the Internet. Of course, this probability is the same that \\ P(\overline{X} < 89.87). Then

\\ P(z

However, we are looking for P(z>1.56), which is the <em>complement probability</em> of the previous probability. Therefore

\\ P(z>1.56) = 1 - P(z

\\ P(z>1.56) = P(\overline{X}>89.87) = 0.0594

Thus, "The probability that the sample mean would be greater than 89.87 WPM" is about \\ P(z>1.56) = 0.0594.

5 0
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Approximately 14 percent of the population of Arizona is 65 years or older. A random sample of five persons from this population
Andreas93 [3]

Answer:

A)0.8533

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are 65 or older, or they are not. The probability of a person being 65 or older is independent of any other person. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

14 percent of the population of Arizona is 65 years or older.

This means that p = 0.14

A random sample of five persons from this population is taken.

This means that n = 5

The probability that less than 2 of the 5 are 65 years or older is :

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.14)^{0}.(0.86)^{5} = 0.4704

P(X = 1) = C_{5,1}.(0.14)^{1}.(0.86)^{4} = 0.3829

P(X < 2) = P(X = 0) + P(X = 1) = 0.4704 + 0.3829 = 0.8533

So the correct answer is:

A)0.8533

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Answer:

Step-by-step explanation

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