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sveta [45]
3 years ago
14

Suppose that 75% of adult Americans agree that, while downloading music from the Internet and then selling it is piracy and shou

ld be prohibited, downloading for personal use is an innocent act and should not be prohibited.We will take a random sample of 50 American adults and ask them if they agree with the statement. Then the sampling distribution of the sample proportion of people who answer yes to the question is:
a approximately Normal, with unknown mean and standard deviation.
b neither Normal, not Binomial.
c Binomial, with n=50 and p=0.75.
d approximately Normal, with mean 0.75 and standard error 0.0612.
Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0
With random sample of 50 and probability of adult American agrees is 75%, then the sampling distribution of the sample proportion is the Binomial distribution, with n=50 and p=0.75

We use Binomial distribution when the number of observation is fixed, each observation is fixed and each observation represents an outcome of "success" and "failure". In above case, the number of observation is 50 and the outcome of success is 0.75
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1 year ago
The mean resting pulse rate for men is 72 beats per minute. A simple random sample of men who regularly work out at Mitch's Gym
BARSIC [14]

Answer:

We conclude that the mean resting pulse rate for men is 72 beats per minute.

Step-by-step explanation:

We are given that he mean resting pulse rate for men is 72 beats per minute. Assume that the standard deviation of the resting pulse rates of all men who work out at Mitch's Gym is known to be 2.6 beats per minute.

A simple random sample of men who regularly work out at Mitch's Gym is obtained and their resting pulse rates (in beats per minute) are listed below;

87, 89, 69, 63, 70, 65, 88, 84, 58, 53, 66, 70.

Let \mu = <u><em>mean resting pulse rate for men.</em></u>

SO, Null Hypothesis, H_0 : \mu = 72 beats/minute     {means that the mean resting pulse rate for men is 72 beats per minute}

Alternate Hypothesis, H_A : \mu < 72 beats/minute      {means that the mean resting pulse rate for men is less than 72 beats per minute}

The test statistics that would be used here <u>One-sample z-test statistics</u> as we know about population standard deviation;

                            T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean mean resting pulse rate = \frac{\sum X}{n}

                =  \frac{87+ 89+ 69 +63 +70 +65+ 88+ 84+ 58+ 53+ 66+ 70}{12}  = 71.83 beats/minute

            \sigma = population standard deviation = 2.6 beats per minute

            n = sample of men = 12

So, <u><em>the test statistics</em></u>  =  \frac{71.83 - 72}{\frac{2.6}{\sqrt{12} } }    

                                     =  -0.23

The value of z test statistics is -0.23.

<u></u>

<u>Also, P-value of the test statistics is given by;</u>

            P(Z < -0.23) = 1 - P(Z \leq 0.23)

                                 = 1 - 0.59095 = <u>0.40905</u>

Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.23 > -1.645, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <em><u>we fail to reject our null hypothesis</u></em>.

Therefore, we conclude that the mean resting pulse rate for men is 72 beats per minute.

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Step-by-step explanation:

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(x + 3) ^{ \frac{1}{2} }  - 1 = x  \\  \sqrt{x  + 3}   - 1 = x \\  \sqrt{x + 3}  = x + 1 \\ square \: both \: sides \\ x + 3 =  {x}^{2}  + 2x + 1 \\  {x}^{2}  + 2x - x + 1 - 3 = 0  \\  {x}^{2}  + x - 2 = 0  \\ x ^{2}  + 2x - x - 2 = 0  \\ (x^{2}  + 2x) - (x  + 2) = 0 \\ x(x + 2) - 1( x + 2) = 0 \\ (x - 1)(x + 2) = 0 \\ x - 1 = 0 \\ x = 1 \\  \\  \\ x  +  2 = 0 \\ x =  - 2

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