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djyliett [7]
3 years ago
10

An automated filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample va

riance of fill volume of s2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid ounces)2, an unacceptable proportion of bottles will be underfilled or overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with underfilled or overfilled bottles?
Chemistry
2 answers:
lions [1.4K]3 years ago
8 0

An automated filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of S² = 0.0153 (fluid ounces)². If the variance of fill volume exceeds 0.01 (fluid ounces)², an unacceptable proportion of bottles will be underfilled or overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with under filled or overfilled bottles? Use α = 0.05, and assume that fill volume has a normal distribution

Answer:

This implies that there is no problems in incorrectly filled because there weren't strong to sustain the there is a problem

Explanation:

Using hypothesis test can come in handy for to measure and accept the real variance and standard deviation of a population distribution

Following the eight steps procedure would guide;

1. Identify the parameter of interest is the population : Variance denoted as S²

2. Formulate the Null hypothesis

H° ;S² = 0.01

3. Draw an alternate hypothesis H¹: S² > 0.01

4.α = 0.05

5. Use Chi ² Test.

The test statistic is:

taurus [48]3 years ago
4 0

Answer:

No, there is no evidence that the manufacturer has a problem with underfilled or overfilled bottles, due that according our results we cannot reject the null hypothesis.

Explanation:

according to this exercise we have the following:

σ^2 =< 0.01 (null hypothesis)

σ^2 > 0.01 (alternative hypothesis)

To solve we can use the chi-square statistical test. To reject or not the hypothesis, we have that the rejection region X^2 > 30.14

Thus:

X^2 = ((n-1) * s^2)/σ^2 = ((20-1)*0.0153)/0.01 = 29.1

Since 29.1 < 30.14, we cannot reject the null hypothesis.

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