A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kg with a standard deviation of 0.5 kg. A random sample of 50 lines were tested and found to have a mean breaking strength of 7.8 kg. With this, does the manufacturer has reasons to believe that the mean breaking strength of the new synthetic fishing line deviates from what he claims?Determine the type of test to use.z testTwo tailed testRight tailed testLeft tailed testt test
Question
A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kg with a standard deviation of 0.5 kg. A random sample of 50 lines were tested and found to have a mean breaking strength of 7.8 kg. With this, does the manufacturer has reasons to believe that the mean breaking strength of the new synthetic fishing line deviates from what he claims?Determine the type of test to use.z testTwo tailed testRight tailed testLeft tailed testt test
Solution
The appropriate test to use in this scenario is the z-test. This is because we know the population standard deviation and the sample size is large (n > 30).
The null hypothesis (H0) is that the mean breaking strength of the new synthetic fishing line is 8 kg, as the manufacturer claims. The alternative hypothesis (H1) is that the mean breaking strength is not 8 kg.
This is a two-tailed test because we are interested in deviations from the claimed mean in either direction, not just whether it is less or greater.
Here are the steps to perform the test:
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Calculate the standard error (SE) of the sample mean. SE = standard deviation / sqrt(n), where n is the sample size. In this case, SE = 0.5 / sqrt(50) = 0.0707 kg.
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Calculate the z-score. This is the difference between the sample mean and the population mean, divided by the standard error. In this case, z = (7.8 - 8) / 0.0707 = -2.83.
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Look up the p-value associated with this z-score. If the p-value is less than the chosen significance level (usually 0.05), then we reject the null hypothesis.
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In this case, a z-score of -2.83 corresponds to a p-value of 0.0047, which is less than 0.05. Therefore, we reject the null hypothesis and conclude that the manufacturer has reason to believe that the mean breaking strength of the new synthetic fishing line deviates from what he claims.
Similar Questions
A manufacturer of sports equipment has developed a new synthetic fishing line that he claims has a mean breaking strength of 8 kg with a standard deviation of 0.5 kg. A random sample of 50 lines were tested and found to have a mean breaking strength of 7.8 kg. With this, does the manufacturer has reasons to believe that the mean breaking strength of the new synthetic fishing line deviates from what he claims?Make a statistical conclusion?There is no enough evidence to say that the claim is accurate.There is no enough evidence to say that the mean breaking strength of the new synthetic fishing line varies from 8 kg.There is enough evidence to say that the mean breaking strength of the new synthetic fishing line varies from 8 kg.There is enough evidence to say that the claim is accurate.
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