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kirill [66]
3 years ago
6

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such tha

t it would produce a mean pressure of 8.0 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 250 engines and the mean pressure was 8.1 pounds/square inch. Assume the variance is known to be 1.00. A level of significance of 0.01 will be used. Make a decision to reject or fail to reject the null hypothesis. Make a decision
Mathematics
1 answer:
disa [49]3 years ago
3 0

Answer:

The calculated value z= 0.0063 does not  fall in the critical region  so we fail to reject the null hypothesis. And accept that the valve does not perform above the specifications.

Step-by-step explanation:

Given that

Population mean= μ = 8.0

Sample mean= x`= 8.1

Sample size= n= 250

Sample standard deviation= s= 1.00

Significance level= ∝ =0.01

The null hypothesis and alternate hypothesis are  

H0: μ < 8.0

Ha: μ ≥ 8.0  One tailed test

The claim is that the valve perform above the specifications

The critical value at 0.01 significance level for one tailed test is z > ±2.33

The test statistic z is used

z= x`- μ/ σ/√n

Z= 8.1-8.0/ 1/√250

z= 0.1/1/15.811

z= 0.0063

The calculated value z= 0.0063  does not fall in the critical region  so we fail to reject the null hypothesis.

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The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
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Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

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Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean:

\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

4\sigma = 56

\sigma = \frac{56}{4} = 14

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

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

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

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

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

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