Question

A house cleaning service claims that it can clean a four bedroom house in less than 2 hours. A sample of n = 16 houses is taken and the sample mean is found to be 1.97 hours and the sample standard deviation is found to be 0.1 hours. Using a 0.05 level of significance the correct conclusion is:(A) reject the null because the test statistic (-1.2) is < the critical value (1.7531).(B) do not reject the null because the test statistic (-1.2) is > the critical value (-1.7531).(C) reject the null because the test statistic (-1.7531) is < the critical value (-1.2).(D) do not reject the null because the test statistic (1.2) is > the critical value (-1.7531).

Answer 1

Answer:

Option B is right

Step-by-step explanation:

$H_0: x bar =2\\H_a: x bar$

(One tailed test at 5% significance level)

n =16 and x bar =1.97

s =0.1

Std error of mean = $\frac{s}{\sqrt{n} } \\=0.025$

Mean diff = $1.97-2 =-0.03$

t statistic =Mean diff/se =-1.2

df =16-1=15

p value =0.124375

(B) do not reject the null because the test statistic (-1.2) is > the critical value (-1.7531).