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just olya [345]
4 years ago
10

In a study of the accuracy of fast food​ drive-through orders, one restaurant had 34 orders that were not accurate among 371 ord

ers observed. Use a 0.01 significance level to test the claim that the rate of inaccurate orders is equal to​ 10%. Does the accuracy rate appear to be​ acceptable? Identify the null and alternative hypotheses for this test?
Mathematics
1 answer:
postnew [5]4 years ago
8 0

Answer:

The claim that he rate of inaccurate orders is equal to​ 10% is supported by statistical evidnece at 5% level

Step-by-step explanation:

Given that in a study of the accuracy of fast food​ drive-through orders, one restaurant had 34 orders that were not accurate among 371 orders observed.

Sample proportion p=0.092\\q=1-p = 0.908\\n = 371

H_0: p =0.10\\H_a: p \neq 0.10

(Two tailed test at 5% significance level)

p difference = 0.092-0.100\\=-0.0084

Std error if H0 is true = \sqrt{\frac{0.1(0.9)}{371} } \\=0.016

Test statistic Z = p diff/std error

=0.539

p value = 0.5899

Since p > 0.05 accept null hypothesis

The claim that he rate of inaccurate orders is equal to​ 10% is supported by statistical evidnece at 5% level

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High blood pressure has been identified as a risk factor for heart attacks and strokes. The proportion of U.S. adults with high
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Answer:

1. It is not appropriate to use the normal curve, since np = 7.4 < 10.

2. The probability that more than 32% of the people in this sample have high blood pressure is 0.0033 = 0.33%.

Step-by-step explanation:

Binomial approximation to the normal:

The binomial approximation to the normal can be used if:

np >= 10 and n(1-p) >= 10

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be 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

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The proportion of U.S. adults with high blood pressure is 0.2. A sample of 37 U.S. adults is chosen.

This means, respectively, that p = 0.2, n = 37

Is it appropriate to use the normal approximation to find the probability that more than 48% of the people in the sample have high blood pressure?

np = 37*0.2 = 7.4 < 10

So not appropriate.

It is not appropriate to use the normal curve, since np = 7.4 < 10.

Part 2:

Now n = 82, 82*0.2 = 16.4 > 10, so ok

Mean and standard deviation:

By the Central Limit Theorem,

Mean \mu = p = 0.2

Standard deviation s = \sqrt{\frac{0.2*0.8}{82}} = 0.0442

Find the probability that more than 32% of the people in this sample have high blood pressure.

This probability is 1 subtracted by the pvalue of Z when X = 0.32. So

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

By the Central Limit Theorem

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

Z = \frac{0.32 - 0.2}{0.0442]

Z = 2.72

Z = 2.72 has a pvalue of 0.9967.

1 - 0.9967 = 0.0033

The probability that more than 32% of the people in this sample have high blood pressure is 0.0033 = 0.33%.

5 0
3 years ago
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