The production of pasture grass for hay is an important agricultural product. A
ID: 3153310 • Letter: T
Question
The production of pasture grass for hay is an important agricultural product. A new type of grass, called Teff Grass is advertised to produce 2 ton of hay per acre on the first cutting. Twenty-five test plots of this new grass resulted in average production of 1.92 ton per acre on the first cutting with a standard deviation of .40 ton. Assume that the production of hay is a normally distributed random variable, which is a necessary assumption with this small sample.
What value must the test statistic be more extreme than in order to reject the null hypothesis at the 1% significance level?
What is the value of the p-value in this case? State a range of values, not a specific value. Hint: The p-value is the tail areas associated with the test statistic, that is the area below the negative and the area above the positive.
What is the decision about the null hypothesis in this case with a significance level of 1%? Hint: If the p -value is less than the significance level then reject the null hypothesis. The p-value is the chance that you are wrong if you reject the null hypothesis based on these data. Write a sentence for the answer.
What is the conclusion about the alternative hypothesis, based on the decision above about the null hypothesis? Write a sentence that begins, Conclude, the data do… .
What is the 99% confidence interval to estimate the parameter based on these data?
Does the above confidence interval contain the hypothesized parameter value in this case? Write a sentence about what the confidence interval indicates about the parameter value.
Explanation / Answer
What value must the test statistic be more extreme than in order to reject the null hypothesis at the 1% significance level?
Formulating the null and alternative hypotheses,
Ho: u = 2
Ha: u =/ 2
As we can see, this is a two tailed test.
Thus, getting the critical t, as alpha = 0.01,
df = n - 1 = 24
tcrit = +/- 2.796939505
Hence, it has to be more extreme than -2.797 or 2.797. [ANSWER]
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What is the value of the p-value in this case? State a range of values, not a specific value. Hint: The p-value is the tail areas associated with the test statistic, that is the area below the negative and the area above the positive.
Getting the test statistic, as
X = sample mean = 1.92
uo = hypothesized mean = 2
n = sample size = 25
s = standard deviation = 0.4
Thus, t = (X - uo) * sqrt(n) / s = -1
Using a table, we see that the P value is
Pvalue > 0.20 [ANSWER]
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What is the decision about the null hypothesis in this case with a significance level of 1%? Hint: If the p -value is less than the significance level then reject the null hypothesis. The p-value is the chance that you are wrong if you reject the null hypothesis based on these data. Write a sentence for the answer.
As P > 0.01, we fail to reject Ho.
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What is the conclusion about the alternative hypothesis, based on the decision above about the null hypothesis? Write a sentence that begins, Conclude, the data do… .
We conclude that there is no significant evidence that the true mean hay per acre is not 2 tons of hay per acre at 0.01 level. [CONCLUSION]
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