: A machine is designed to cometer sticks to be 1000 mm, long 19 meter sticks are measured, and it's found that they have a mean length of 1001.8mm, with a standard deviation of 2.6 mm. To a 1% level of significance, Can it be asserted that the Mandard deviation of the process in general (ie, of all the metersticks produced by the machine) is over 2 min.! I Acume that the lengths of all the meter sticks are normally distributed] (D) p. 2 92 male and 103 female Neptunians are weighed. It's found the males have a mean weight of 422 lbs., with a standard deviation of 30 lbs., while the females have a mean weight of 409 lbs., with a standard deviation of 28 lbs. Toa 1 level of significance, can it be asserted that, in general, male Neptunians weigh more than female Neptunians?

We are testing if the population standard deviation is greater than 2 mm. Since the sample size is small (n < 30) and the population standard deviation is unknown, we should use the chi-squared distribution.

Calculate the chi-squared test statistic:

χ² = ((n - 1) * s²) / σ² = ((19 - 1) * 2.6²) / 2² ≈ 53.947

Using a chi-squared table or calculator, find the critical value for a 1% level of significance with 18 degrees of freedom (n - 1). The critical value is approximately 37.566.

Since the calculated test statistic (χ² ≈ 53.947) is greater than the critical value (37.566), we reject the null hypothesis.

Therefore, at a 1% level of significance, we can assert that the standard deviation of the process is over 2 mm.

Neptunian Weights:

Sample mean for males = 422 lbs.

Sample mean for females = 409 lbs.

Sample standard deviation (s) for males = 30 lbs.

Sample standard deviation (s) for females = 28 lbs.

Sample size for males (n1) = 92

Sample size for females (n2) = 103

Hypotheses:

Null hypothesis (H0): μ1 - μ2 ≤ 0 (No significant difference)

Alternative hypothesis (Ha): μ1 - μ2 > 0 (Males weigh more than females)

We are testing if male Neptunians weigh more than female Neptunians. Since we have two independent samples and population standard deviations are unknown, we should use the t-distribution.

Calculate the test statistic:

t = (mean1 - mean2) / √((s1²/n1) + (s2²/n2)) = (422 - 409) / √((30²/92) + (28²/103)) ≈ 2.961

Using a t-distribution table or calculator with appropriate degrees of freedom, find the critical t-value for a 1% level of significance (1-tailed). The critical value for 1% significance with degrees of freedom around 190 is approximately 2.613.

Since the calculated test statistic (t ≈ 2.961) is greater than the critical value (2.613), we reject the null hypothesis.

Therefore, at a 1% level of significance, we can assert that, in general, male Neptunians weigh more than female Neptunians.

Learn more about hypotheses on:

https://brainly.com/question/606806

#SPJ4