Question

1. Your team has developed a new process that the team thinks will make the company more profit per item. Under the original process, 50 components were manufactured with a net profit per component of $1,255 with a standard deviation of $215. Under the new process, 30 components were manufactured with a net profit of $1,330 with a standard deviation of $238. Calculate the 95% confidence interval of both processes together. Note: Enter the +/- portion into the answer block. As an example: If your 95% confidence interval is 200 +/- 35.3, enter 35.3 into the answer box. Give your answer to two decimal places.

2. A graduate program has 100 applicants. The applicants submit scores on a standardized test as part of their admission package. The test would be of little use to the admissions committee if there was no variance in the scores (everyone made basically the same score on the test). The committee found that the variance of the scores was 127. Calculate the lower limit value of the two-sided 95% confidence interval for variance. Give your answer to two decimal places.

Answer 1

Sorry as per the Chegg guidelines I'm only allowed to solve the first question

1)

Mean1 $(\bar{X_1})$ = 1255

Sample size1 (n1) = 50

Standard deviation1 (s1) = 215

Mean2 $(\bar{X_2})$ = 1330

Sample size2 (n2) = 30

Standard deviation2 (s2) = 238

Confidence interval(in %) = 95

t_{\alpha/2, n_1 + n_2 -2} = 1.99

Since we know that

$\\Confidence\; interval = \bar{X_1}-\bar{X_2} \pm t_{\alpha/2, n-1}S_P\sqrt{\frac{1}{n_1}+\frac{1}{n_2}} \\S_P = \sqrt{\frac{(n_1-1)s1^2 + (n_2-1)s2^2}{n_1 + n_2 - 2}} \\Required\; confidence\; interval = (1255.0-1330.0-1.99S_P\sqrt{\frac{1}{50}+\frac{1}{30}}, 1255.0-1330.0+1.99S_P\sqrt{\frac{1}{50}+\frac{1}{30}})$

Required confidence interval = (-75.0-102.8646, -75.0+102.8646)

Required confidence interval = (-177.86, 27.86)

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