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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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A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.4-cm and a standard dev
Kipish [7]

Answer:

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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In this problem, we have that:

\mu = 111.4, \sigma = 0.5, n = 23, s = \frac{0.5}{\sqrt{23}} = 0.1043

Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.

This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So

X = 111.4

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{111.4 - 111.4}{0.1043}

Z = 0

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X = 111.2

Z = \frac{X - \mu}{s}

Z = \frac{111.2 - 111.4}{0.1043}

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Z = -1.92 has a pvalue of 0.0274.

0.5 - 0.0274 = 0.4726

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