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inysia [295]
3 years ago
7

In Exercises 1–4, use these results from a USA Today survey in which 510 people chose to respond to this question that was poste

d on the USA Today website: "Should Americans replace passwords with biometric security (fingerprints, etc)?" Among the respondents, 53% said "yes." We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.Number and Proportiona. Identify the actual number of respondents who answered "yes."b. Identify the sample proportion and the symbol used to represent it.
Mathematics
1 answer:
Alexus [3.1K]3 years ago
7 0

Answer:

The proportion of the population believes that passwords should be replaced with biometric security is not more than 50%.

Step-by-step explanation:

In this case we need to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Assume that the significance level of the test is, <em>α</em> = 0.05.

The hypothesis can be defined as follows:  

<em>H</em>₀: The proportion of the population believes that passwords should be replaced with biometric security is 50%, i.e. <em>p</em> = 0.50.  

<em>Hₐ</em>: The proportion of the population believes that passwords should be replaced with biometric security is more than 50%, i.e. <em>p</em> > 0.50.  

The information provided is:

n = 510\\\hat p=0.53

Compute the test statistic value as follows:

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

    =\frac{0.53-0.50}{\sqrt{\frac{0.50(1-0.50)}{510}}}\\\\=1.354991\\\\\approx 1.35

The test statistic value is 1.35.

The decision rule is:

The null hypothesis will be rejected if the p-value of the test is less than the significance level.

Compute the p-value as follows:

 p-value=P(Z>1.35)

                 =1-P(Z

<em>p</em>-value = 0.088 > <em>α</em> = 0.05.

The null hypothesis will not be rejected at 5% significance level.

Thus, there is not enough evidence to support the claim.

Conclusion:

The proportion of the population believes that passwords should be replaced with biometric security is not more than 50%.

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Answer:

\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

Step-by-step explanation:

Let us find out the mean savings for a televisit to the doctor from the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

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Let us find out the standard deviation of savings for a televisit to the doctor from the given data.

Using Excel,

=STDEV(number1, number2,....)

The standard deviation is found to be

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The confidence interval is given by

\text {confidence interval} = \bar{x} \pm MoE\\\\

Where the margin of error is given by

$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\

Where n is the sample of 20 online doctor visits, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 95% confidence level.

The t-score is given by is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

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MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\MoE = 2.093\cdot \frac{22.35}{\sqrt{20} } \\\\MoE = 2.093\cdot 4.997\\\\MoE = 10.46\\\\

So the required 95% confidence interval is

\text {CI} = \bar{x} \pm MoE\\\\\text {CI} = 71 \pm 10.46\\\\\text {CI} = 71 - 10.46, \: 71 + 10.46\\\\\text {CI} = (60.54, \: 81.46)\\\\

Therefore, we are 95% confident that actual mean savings for a televisit to the doctor is within the interval of ($60.54 to $81.46)

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Answer: 40 %

Step-by-step explanation:

Let the amount of one card = x

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And, the value of 22 cards ( Total amount after 3 month ) = 22x

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Hence, the annual rate of interest = 40%

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