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Ilya [14]
3 years ago
14

A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us

ing the visual analog scale. Assume that both samples are independent simple random samples from populations having normal distributions. Use a 0.05 significance level to test the claim that those given a sham treatment have pain reductions that vary more than the pain reductions for those treated with magnets.
Sham: n=20, xÌ=0.44,s=1.24n=20, xÌ=0.44,s=1.24
Magnet: n=20, xÌ=0.49,s=0.95

Required:
a. Identify the test statistic. (Round to two decimal places as needed)
b. Use technology to identify the P-value. (Round to three decimal places as needed)
c. What is the conclusion for this hypothesis test?
Mathematics
1 answer:
amm18123 years ago
3 0

Answer:

Given:

Sham: n= 20,     x=0.44,   s=1.24,

Magnet:n= 20,  x =0.49,   s= 0.95

For Sham:

Sample size, n = 20

Sample mean = 0.44

Standard deviation = 1.24

For Magnet:

Sample size = 20

Sample mean = 0.49

Standard deviation = 0.95

The null and alternative hypotheses:

H0: s1²=s2²

H1: s1² ≠ s2²

a) To find the test statistics, use the formula:

\frac{s1^2}{s2^2}

\frac{1.24^2}{0.95^2} = \frac{1.5376}{0.9025} = 1.7037

Test statistics = 1.7037

b) P-value:

Sham: degrees of freedom = n - 1 = 20 - 1 = 19

Magnet: degrees of freedom = n - 1 = 20 - 1 = 19

The critical values:

[Za/2, df1, df2)], [(1 - Za/2), df1, df2]

f[0.05/2, 19, 19], f[(1 - 0.05/2), 19, 19]

f[0.025, 19, 19], f[0.975, 19, 19]

(2.526, 0.3958)

The rejection region:

Reject H0, if  F < 0.3958 or if F > 2.526

c) Conclusion:

Since the critical values of test statistic is between (0.3958 < 1.7037 < 2.526), we fail to reject null hypothesis H0.

There is insufficient evidence to to support the claim that those given a sham treatment have reductions that vary more than those treated with magnets

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A highway traffic condition during a blizzard is hazardous. Suppose one traffic accident is expected to occur in each 60 miles o
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Answer:

a) 0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

b) 0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

c) 0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

Step-by-step explanation:

We have the mean for a distance, which means that the Poisson distribution is used to solve this question. For item b, the binomial distribution is used, as for each blizzard day, the probability of an accident will be the same.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

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.

Suppose one traffic accident is expected to occur in each 60 miles of highway blizzard day.

This means that \mu = \frac{n}{60}, in which 60 is the number of miles.

(a) What is the probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway?

n = 25, and thus, \mu = \frac{25}{60} = 0.4167

This probability is:

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.4167}*(0.4167)^{0}}{(0)!} = 0.6592

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6592 = 0.3408

0.3408 = 34.08% probability that at least one accident will occur on a blizzard day on a 25 mile long stretch of highway.

(b) Suppose there are six blizzard days this winter. What is the probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway?

Binomial distribution.

6 blizzard days means that n = 6

Each of these days, 0.6592 probability of no accident on this stretch, which means that p = 0.6592.

This probability is P(X = 2). So

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

P(X = 2) = C_{6,2}.(0.6592)^{2}.(0.3408)^{4} = 0.0879

0.0879 = 8.79% probability that two out of these six blizzard days have no accident on a 25 mile long stretch of highway.

(c) If the probability of damage requiring an insurance claim per accident is 60%, what is the probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway?

Probability of damage requiring insurance claim per accident is of 60%, which means that the mean is now:

\mu = 0.6\frac{n}{60} = 0.01n

80 miles:

This means that n = 80. So

\mu = 0.01(80) = 0.8

The probability of no damaging accidents is P(X = 0). So

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

0.4493 = 44.93% probability of no damaging accidents requiring a insurance claims on an 80 mile stretch of highway.

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In which:

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The margin of error is given by:

M = z\frac{\sigma}{\sqrt{n}}

In this problem, we have that the parameters are given as follows:

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Hence, solving for n, we find the sample size.

M = z\frac{\sigma}{\sqrt{n}}

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3\sqrt{n} = 1.645 \times 16

\sqrt{n} = \frac{1.645 \times 16}{3}

(\sqrt{n})^2 = \left(\frac{1.645 \times 16}{3}\right)^2

n = 76.97

Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.

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