Math  /  Data & Statistics

QuestionEyeglassomatic manufactures eyeglasses for different retailers. They test to see how many defective lenses they made in a given time period and found that 12%12 \% of all lenses had defects of some type. Looking at the type of defects, they found in a three-month time period that out of 34,054 defective lenses, 4257 were due to scratches. Are there more defects from scratches than from all other causes? Use a 1%1 \% level of significance.
State the hypotheses. H0:pH_{0}: p ? \square Ha:p?H_{a}: p ? \square Calculate the test statistic. Round to four decimal places. p^=\hat{p}= \square Calculate the standardized test statistic. Round to three decimal places. z=z=\square
Find the pp-value. Round to four decimal places. pp-value == \square State your decision. Since the pp-value is greater than .01 , fail to reject H0H_{0}. Since the pp-value is greater than .01 , reject H0H_{0}. Since the pp-value is less than .01 , reject H0H_{0}. Since the pp-value is less than .01 , fail to reject H0H_{0}.
Interpret the results. At the 1%1 \% level of significance, there is enough evidence to show that the proportion of defective lenses from scratches is lower than 12%12 \%, the proportion from other causes. At the 1%1 \% level of significance, there is enough evidence to show that the proportion of defective lenses from scratches is higher than 12%12 \%, the proportion from other causes. At the 1%1 \% level of significance, there is not enough evidence to show that the proportion of defective lenses from scratches is lower than 12%12 \%, the proportion from other causes. At the 1%1 \% level of significance, there is not enough evidence to show that the proportion of defective lenses from scratches is higher than 12%12 \%, the proportion from other causes. At the 1%1 \% level of significance, there is not enough evidence to show that the proportion of defective lenses from scratches is not equal to 12%12 \%, the proportion from other causes.

Studdy Solution

STEP 1

1. We are dealing with a proportion hypothesis test.
2. The significance level is α=0.01 \alpha = 0.01 .
3. The sample size is 34,054 defective lenses.
4. The number of defective lenses due to scratches is 4,257.

STEP 2

1. State the null and alternative hypotheses.
2. Calculate the sample proportion.
3. Calculate the standardized test statistic.
4. Find the p-value.
5. Make a decision based on the p-value.
6. Interpret the results.

STEP 3

State the null and alternative hypotheses:
- H0:p=0.12 H_0: p = 0.12 (The proportion of defects due to scratches is equal to 12%) - Ha:p>0.12 H_a: p > 0.12 (The proportion of defects due to scratches is greater than 12%)

STEP 4

Calculate the sample proportion p^\hat{p}:
p^=4257340540.1250\hat{p} = \frac{4257}{34054} \approx 0.1250

STEP 5

Calculate the standardized test statistic z z :
z=p^p0p0(1p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
Where p0=0.12 p_0 = 0.12 and n=34054 n = 34054 .
z=0.12500.120.12×0.88340542.820z = \frac{0.1250 - 0.12}{\sqrt{\frac{0.12 \times 0.88}{34054}}} \approx 2.820

STEP 6

Find the p-value for the test statistic z=2.820 z = 2.820 .
Using a standard normal distribution table or calculator:
p-value0.0024p\text{-value} \approx 0.0024

STEP 7

Make a decision based on the p-value:
Since the p-value=0.0024 p\text{-value} = 0.0024 is less than 0.01 0.01 , we reject H0 H_0 .

STEP 8

Interpret the results:
At the 1% 1\% level of significance, there is enough evidence to show that the proportion of defective lenses from scratches is higher than 12% 12\% , the proportion from other causes.

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