Question

The company’s cleaning service states that they spend more than 46 minutes each time the cleaning service is there. The company times the length of 37 randomly selected cleaning visits and finds the average is 48.2 minutes. Assuming a population standard deviation of 5.2 minutes, can the company support the cleaning service’s claim at α=0.10?

Answer 1

Answer:

Step-by-step explanation:

Let X be time that the company cleaning service spends for cleaning

Given that X bar = sample mean = 48.2 min.  n = 37

sigma = 5.2 minutes

Std error of sample = $\frac{5.2}{\sqrt{37} } =0.8549$

$H_0: mu = 46\\H_a:mu >46$

(One tailed test)

Mean difference = $46-48.2 = -2.2$

Test statistic = Mean diff/se = $\frac{-2.2}{0.8549} =-2.573$

p value = 0.0041

Since p <0.10 we reject H0.  The average time is greater than 46 minutes.