Question

The city of Providence supervisor claims that the mean weight of all garbage discarded by households each week is 31 lb. A random sample of 49 households shows that the sample mean is 30.14 with a standard deviation of 10.06 lb. Is there evidence to reject the supervisor's claim? (use α=0.05 and the rejection region approach).

Answer 1

Answer:

$\mu$= 31, n = 49,  $\bar x$= 30.14, s= 10.06, $\alpha$ = 0.05

Ho: $\mu$ = 31

Ha: $\mu$ $\neq$ 31

Calculate test statistics

- s/n

30.14 - 31 10.06/149

t = -0.598

test statistics = -0.598

now calculate t critical value to get rejection region

find critical t value using t table with $\alpha$ = 0.05 and df = n-1 = 48

we get

t critical value = ( -2.011 , 2.011)

Decision rule is

if (test statistics < -2.011) and (if (test statistics > 2.011)

then Reject Ho.

here

( -0.598 > -2.011) and ( 0.598 <  2.011)

Thus our test statistics do not lie in rejection region

Hence  It is concluded that the null hypothesis Ho is not rejected.

Therefore, there is not enough sufficient evidence to reject the supervisor's claim