A new process has been designed to make ceramic tiles. The goal is to have no more than 5% of the tiles be nonconforming due to surface defects. A random sample of 1000 tiles is inspected. Let p̂ be the proportion of nonconforming tiles in the sample. a) If the sample proportion of nonconforming tiles were 0.075, would it be pausible that the goal had been reached? Explain.b) If 5% of the tiles produced are nonconforming, what is P (p̂ ≥ 0.075)? c) Based on the answer to part a, if 5% of the tiles are nonconforming, is a proportion of 0.075 nonconforming tiles in a sample of 1000 unusually large?

Question
Asked Apr 30, 2019

A new process has been designed to make ceramic tiles. The goal is to have no more than 5% of the tiles be nonconforming due to surface defects. A random sample of 1000 tiles is inspected. Let p̂ be the proportion of nonconforming tiles in the sample. 

a) If the sample proportion of nonconforming tiles were 0.075, would it be pausible that the goal had been reached? Explain.

b) If 5% of the tiles produced are nonconforming, what is P (p̂ ≥ 0.075)? 

c) Based on the answer to part a, if 5% of the tiles are nonconforming, is a proportion of 0.075 nonconforming tiles in a sample of 1000 unusually large? 

check_circleExpert Solution
Step 1

Part a:

It can be checked whether the goal of “not more than 5% non-conforming tiles” is reached in case of sample proportion of 0.075, by testing the following hypotheses:

Null hypothesis:

H0: p = 0.05, that is, the sample of 1,000 tiles does not have more than 5% non-conforming tiles.

Alternative hypothesis:

H0: p > 0.05, that is, the sample of 1,000 tiles has more than 5% non-conforming tiles.

The sampling distribution of , using Central Limit Theorem for large samples, is Normal distribution, with mean, μ = p = 0.05 and standard deviation, σ = √[p (1 – p)/n] = √[0.05 ∙ (1 – 0.05)/1,000] ≈ 0.0069.

Let the level of significance be α = 0.01.

The p-value of the test, that is, the probability that p takes a value at least as extreme as = 0.075, given that the null hypothesis is true, that is, given that the true value is p = 0.05, is:

P (≥ 0.075| H0 is true) = P (≥ 0.075| p = 0.05) = 1 – P (< 0.075| p = 0.05) = 1 – 0.9999 [Use EXCEL formula: =NORM.DIST(0.075,0.05,0.0069,1)] = 0.0001.

The rule regarding rejection of the hypothesis is: “Reject H0 if p-value < α”.

Here, p-value (= 0.0001) < α (= 0.01). Thus, there is enough evidence to reject H0.

Hence, it is not plausible that the goal is reached.

Step 2

Part b:

It is found from part a that P (≥ 0.075) when H0 is true, that is, when 5% of the tile produced are nonconforming, is 0.0001.

Step 3

Part c:

If the value of a normally distributed random variable is more than 3 standard deviations away from the mean, then it is considered as unusual.

Now, 3 standard deviations above the mean is:

μ + 3σ = p + 3√[p (1 – p)/n] = 0.05 + (3 ∙ 0.0069...

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Hypothesis Testing

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