Question

an agronomist examines the cellulose content if alfalfa gay. suppose that the cellulose content in the population had a standard deviation of 8 mg/g. a sample of 15 cuttings has a mean cellulose of 145mg/g.

a. give a 90% confidence interval for the true population mean cellulose content

b. a previous study claimed that the mean cellulose content was 140mg/g, but the agronomist has a reason to believe that the mean us higher than the figure. State the hypotheses and carry out a significance test to see if the new data support this belief

c. what assumptions do you need to make for a statistical procedures to be valid?.

Answer 1

Here, we have given that,

Xi: cellulose mg/g

n= Number of cuttings = 15

= sample mean cellulose = 145 mg/g     

= Population standard deviation=8 mg/g

(A)

Now, we want to find the 90% confidence interval for the true population mean cellulose content  

The formula is as follows,

Where

E=Margin of error =

Now, first, we can find the critical value,

c=confidence level =0.90

=level of significance=1-c=1-0.90=0.10

and we know that confidence interval is always two-tailed

                             =      Using EXCEL software= ABS(NORMSINV(probablity=0.05))

                             = 1.645

Now,

E=

   =

= 3.3979

We get the 90% confidence interval for true population mean cellulose content  

Interpretation:

We conclude that we are 90% confident that the population mean will fall within this confidence interval

(B)

Claim: To check whether the population mean cellulose content is higher than 140 mg/g.

The null and alternative hypothesis is as follows

mg/g

Versus

mg/g

where, = population mean cellulose content

This is the right one-tailed test

Now, we can find the test statistic

z-statistics =

                  =

                  =

                   =2.42

The Test statistic is 2.42

Now we can find the P-value

P-value = P(Z > z-statistics) as this is the right one-tailed test

               =1- P( Z < 2.42)

                =1 - 0.99224 Using standard normal z table see the value corresponding to the z=2.42

= 0.0078

we get the P-value is 0.0078

Decision:

c=confidence level =0.90

=level of significance=1-c=1-0.90=0.10

Here, P-value (0.0078) less than (<) 0.05 ( )

Conclusion

we reject Ho (Null Hypothesis)

There is sufficient evidence to support the claim the population mean cellulose content is higher than 140 mg/g.

(C)

The below mentioned necessary assumption is satisfied for constructing the statistical hypothesis test and 90% Confidence Interval (one sample z-test).

• The data are a simple random sample as each observation are independent of each other.
• Distribution of the data is assumed to be normal.
• The population standard deviation is known we are using the one-sample z-test to test the hypothesis.