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Sergio [31]
3 years ago
10

When working properly, a machine that is used to makes chips for calculators does not produce more than 4% defective chips. When

ever the machine produces more than 4% defective chips, it needs an adjustment. To check if the machine is working properly, the quality control department at the company often takes samples of chips and inspects them to determine if they are good or defective. One such random sample of 200 chips taken recently from the production line contained 12 defective chips. Find the p-value to test the hypothesis whether or not the machine needs an adjustment. What would your conclusion be if the significance level is 2.5%
Mathematics
1 answer:
Anika [276]3 years ago
8 0

Answer:

The proportion of defective chips produced by the machine is more than 4% so the machine needs an adjustment.

Step-by-step explanation:

In this case we need to test whether the proportion of defective chips produced by the machine is more than 4%.

The hypothesis can be defined as follows:

<em>H</em>₀: The proportion of defective chips produced by the machine is not more than 4%, i.e. <em>p</em> ≤ 0.04.

<em>H</em>ₐ: The proportion of defective chips produced by the machine is more than 4%, i.e. <em>p</em> > 0.04.

The information provided is:

<em>X</em> = 12

<em>n</em> = 200

<em>α</em> = 0.025

The sample proportion of defective chips is:

\hat p=\frac{X}{n}\\\\=\frac{12}{200}\\\\=0.06

Compute the test statistic as follows:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\\\\=\frac{0.06-0.04}{\sqrt{\frac{0.04(1-0.04)}{200}}}\\\\=1.44

The test statistic value is 1.44.

Decision rule:

We reject a hypothesis if the p-value of a statistic is lower than the level of significance α.

Compute the <em>p</em>-value of the test:

p-value=P(Z>1.44)\\=1-P(Z

The <em>p</em>-value of the test is 0.075.

<em>p</em>-value = 0.075 > <em>α</em> = 0.025

The null hypothesis was failed to be rejected at 2.5% level of significance.

Thus, it can be concluded that the proportion of defective chips produced by the machine is more than 4% so the machine needs an adjustment.

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Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples 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.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

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

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

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

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

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

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

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