Question

Ms. Fitness-Buff, a high school gym teacher, wants to propose an after-school fitness program. To get an idea of the fitness level of the students at her school, she takes a random sample of 75 students and records the number of hours the students exercised in the past week. Her sample mean is 2.25 hours and she knows from past research that the population standard deviation is 2 hours. She wants to know if this varies from a population mean of 3 hours/week.
1 .Construct a 95% confidence interval
2. Draw a conclusion for a two-sided test at a = .05.

Answer 1

Answer:

a) Confidence interval 95% = Mean ±Margin of error

= $(-1.7974, 2.7026)$

b) since 3 is not contained in the interval, we reject null hypothesis.

Step-by-step explanation:

Given that Ms. Fitness-Buff, a high school gym teacher, wants to propose an after-school fitness program. To get an idea of the fitness level of the students at her school, she takes a random sample of 75 students and records the number of hours the students exercised in the past week. Her sample mean is 2.25 hours and she knows from past research that the population standard deviation is 2 hours.

She wants to know if this varies from a population mean of 3 hours/week.

$H_0: \bar x = 3\\H_a: \bar x\neq 3$

(Two tailed test at 5% significance level)

Sample mean = 2.25 hours

Since sigma is known, we use Z critical value for 95% i.e. 1.96

Margin of error = $1.96*\frac{2}{\sqrt{75} } \\=0.4526$

a) Confidence interval 95% = Mean ±Margin of error

= $(-1.7974, 2.7026)$

b) since 3 is not contained in the interval, we reject null hypothesis.