A random sample of 43 cows was selected to investigate the claim that the mean weekly milk yield for cows is greater than 120 kilograms. The sample mean weekly milk yield is 127.8 and the sample standard deviation is 22.8. Let μ be the population mean weekly milk yield for cows.
(give answer to TWO places past decimal)
1. Construct a 90% lower confidence bound for μ.
Lower Bound:
| Tries 0/99 |
2. Perform a hypothesis test to see if the claim is true (HA : μ > 120) at the significant level α = 0.05:
Compute the test statistic.
| Tries 0/99 |
Indicate the rejection region:
z < z0.05
z > z0.025
z < z0.025
z > z0.05
| Tries 0/99 |
Is there enough evidence to reject the Null hypothesis: H0 : μ =
120 ?
No, don't reject.
Yes, reject.
| Tries 0/99 |
Ans:
1)Lower confidence bound=127.8-1.282*(22.8/SQRT(43))=123.34
2)
Test statistic:
z=(127.8-120)/(22.8/SQRT(43))
z=2.24
rejection region: z>z0.05
(z>1.645)
3)critical z value=1.645
As,test statistic z>1.645,so
Yes,reject H0.
A random sample of 43 cows was selected to investigate the claim that the mean weekly...