Question

4. Sarah is interested in the average level of "openness" of the 50 students in her USC Statistics class, on a scale of 1 to 10. She takes a simple random sample of 20 students in her class, interviews them, and finds the average level of openness to be 6, with a standard deviation of 1.5. Find a 98% confidence interval for the average level of openness for all students in her class

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Answer 1

98% confidence interval for is

98% confidence interval for mu is x^bar- t * S/ squareroot (n) < mu < x^bar + t * S/ squareroot (n) 6 - 2.539 * 1.5/sqrt(20) < mu < 6 + 2.539 * 1.5/sqrt(20) 5.1484 < mu < 6.8516 98% CI for mu is (5.1484 , 6.8516)

$\bar{x}$ - t * S / sqrt ( n) < < $\bar{x}$ + t * S / sqrt ( n)

98% confidence interval for mu is x^bar- t * S/ squareroot (n) < mu < x^bar + t * S/ squareroot (n) 6 - 2.539 * 1.5/sqrt(20) < mu < 6 + 2.539 * 1.5/sqrt(20) 5.1484 < mu < 6.8516 98% CI for mu is (5.1484 , 6.8516)

6 - 2.539 * 1.5 / sqrt(20) < < 6 + 2.539 * 1.5 / sqrt(20)

98% confidence interval for mu is x^bar- t * S/ squareroot (n) < mu < x^bar + t * S/ squareroot (n) 6 - 2.539 * 1.5/sqrt(20) < mu < 6 + 2.539 * 1.5/sqrt(20) 5.1484 < mu < 6.8516 98% CI for mu is (5.1484 , 6.8516)

5.1484 < < 6.8516

98% confidence interval for mu is x^bar- t * S/ squareroot (n) < mu < x^bar + t * S/ squareroot (n) 6 - 2.539 * 1.5/sqrt(20) < mu < 6 + 2.539 * 1.5/sqrt(20) 5.1484 < mu < 6.8516 98% CI for mu is (5.1484 , 6.8516)

98% CI for is (5.1484 , 6.8516)

98% confidence interval for mu is x^bar- t * S/ squareroot (n) < mu < x^bar + t * S/ squareroot (n) 6 - 2.539 * 1.5/sqrt(20) < mu < 6 + 2.539 * 1.5/sqrt(20) 5.1484 < mu < 6.8516 98% CI for mu is (5.1484 , 6.8516)