Answer:
H0: μ ≥ 16
H1: < 16
−2.748749;
0.011275;
Reject the Null
Step-by-step explanation:
Given the data:
15.87, 16.02, 15.78, 15.83, 15.69, 15.81, 16.04, 15.81, 15.92, 16.10
Null hypothesis ; H0: μ ≥ 16
Alternative hypothesis ; H1: < 16
Sample size, n = 10
From the data:
Using calculator,
Sample mean, m = 15.887
Standard deviation, s = 0.13
The test statistic, T
(m - μ) / s/sqrt(n)
(15.887 - 16) / (0.13/sqrt(10))
= −2.748749
Using the p value from test statistic calculator :
Degree of freedom (df) = 10 - 1 = 9 at 0.05 significance level is 0.011275
Since the p value is < 0.05
0.011 < 0.05
We reject the Null and conclude that the mean fill weight is less than 16 oz
