STATE: What is the question of interest? A) is the sample mean for the pepper pl
ID: 3174914 • Letter: S
Question
STATE: What is the question of interest? A) is the sample mean for the pepper plant less than 80 days? B) is the true mean different than 80? C) is there evidence the true mean time to maturity is greater than 80 days? Use the following for the next 5 questions: The time lin number of days) until maturity of a certain variety of hot pepper is Normally distributed. with unknown mean Hand standard deviation a 2.4. This variety is advertised as taking 80 days to mature. Suppose you wish to test whether the number of days to mature is less than 8o days if you use a certain growing regimen. You select a simple random sample of four plants of this variety and measure the time until maturity. The four times, in days, are 76 78 71 81 Use a significance level of 0.05.Explanation / Answer
Solution1:
Null hypothesis
Ho: mean,mu=80
Alternatiove Hypothesis:
H1:mean,mu<80
one tail test
level of significance=0.05
test statistic:
t=samplemean-populationmean/population std dev/sqrt(sampleszie)
=(76+78+71+81/4)-80/2.4sqrt(4)
=76.5-80/2.4/2
=-3.5/1.2
t =-2.92
p value for alpha=0.05 and degrees of freedom=n-1=4-1=3
The P-Value is .030748.
0.030748<0.05which is level of signiifcance
The result is significant at p < .05.
Reject Nulll hypothesis
Accept alternative hypothesis
conclusion;
there is sufficient evidence at 5% level of significance to support the claim that
he sample mean for the pepper plant less than 80 days
Solution:
Null hypothesis:
Ho:mean=80
Alternative Hypothesis:
H1:mean not equal to 80
two tail test
level of sdigniifcance=0.05
test statistic remain same
t=-2.92
The P-Value is .061496.
0.0614>0.05
The result is not significant at p < .05.
Fail to reject Null hypothesis
Accept Null hypothesis.
Conclusion:
there is insufficient evidence at 5% level of significance to support the claim that
he true mean different than 80
SOlutionc:
Null hypothesis:
Ho:mean=80
alternative Hypothesis:
H1::mean>80
level of significance=0.05
t=-2.92
The P-Value is .030748.
The result is significant at p < .05.
Reject null hypothesis:
Accept Alternatove Hypothesis.
there is sufficent evidence at 5% level of significance to support the claim
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