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QveST [7]
3 years ago
12

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 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.
n xbar s
Sham 20 0.41 1.26
Magnet 20 0.46 0.93
Mathematics
1 answer:
gregori [183]3 years ago
4 0

Answer and Step-by-step explanation: The null and alternative hypothesis for this test are:

H_{0}: s_{1}^{2} = s_{2}^{2}

H_{a}: s_{1}^{2} > s_{2}^{2}

To test it, use F-test statistics and compare variances of each treatment.

Calculate F-value:

F=\frac{s^{2}_{1}}{s^{2}_{2}}

F=\frac{1.26^{2}}{0.93^{2}}

F=\frac{1.5876}{0.8649}

F = 1.8356

The <u>critical value of F</u> is given by a F-distribution table with:

degree of freedom (row): 20 - 1 = 19

degree of freedom (column): 20 - 1 = 19

And a significance level: α = 0.05

F_{critical} = 2.2341

Comparing both values of F:

1.856 < 2.2341

i.e. F-value calculated is less than F-value of the table.

Therefore, failed to reject H_{0}, meaning there is <u>no sufficient data to support the claim</u> that sham treatment have pain reductions which vary more than for those using magnets treatment.

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Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age with a mean of 93 seconds and
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Answer: The running time should at least 119.32 seconds to be in the top 5% of runners.

Step-by-step explanation:

Let X= random variable that represents the running time of men between 18 and 30 years of age.

As per given, X is normally distrusted with mean \mu=93\text{ seconds} and standard deviation \sigma=16\text{ seconds}.

To find: x in top 5% i.e. we need to find x such that P(X<x)=95% or 0.95.

i.e. P(\dfrac{X-\mu}{\sigma}

P(Z

Since, z-value for 0.95 p-value ( one-tailed) =1.645

So,

Hence, the running time should at least 119.32 seconds to be in the top 5% of runners.

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Find the surface area and volume of a cube with sides that are 6 inches.
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8 0
3 years ago
Test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.
VikaD [51]

Answer:

1. H0 : p = 0.9

   H1 : p ≠ 0.9

2. The test is two tailed.

3. Reject the null hypothesis

Step-by-step explanation:

We are given that we have to test the claim that the proportion of people who own cats is significantly different than 90% at the 0.1 significance level.

So, Null Hypothesis, H_0 : p = 0.90

Alternate Hypothesis, H_1 : p \neq 0.9

Here, the test is two tailed because we have given that to test  the claim that the proportion of people who own cats is significantly different than 90% which means it can be less than 0.90 or more than 0.90.

Now, test statistics is given by;

            \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)   , where,  n = sample size = 100

                                                       \hat p = 0.94 (given)

So, Test statistics = \frac{0.94-0.90}{\sqrt{\frac{0.94(1-0.94)}{100} } } = 1.68

Now, P-value = P(Z > 1.68) = 1 - P(Z <= 1.68)

                                           = 1 - 0.95352 = 0.0465 ≈ 0.05 or 5%

Now, our decision rule is that;

       If p-value < significance level  ⇒ Reject null hypothesis

       If p-value > significance level  ⇒ Accept null hypothesis

Since, here p-value is less than significance level as 0.05 < 0.1, so we have sufficient evidence to reject null hypothesis.

Therefore, we conclude that proportion of people who own cats is significantly different than 90%.

8 0
3 years ago
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