4. A fertilize company tests blends of fertilizer with hybrid seeds to accurately determine yields. Consistency is desired, so the fertilizer company wants to test the variability of yields. In years past, the farm harvested an average of 88 bushels per acre, with a standard deviation of 6 bushels per acre. acre plots the fertilize company collected data suggesting that the standard deviation of the yield was 5 bushels per acre. blend higher than the old? This year, the farm altered the blend a bit. On 20 one- Is the variance of the new The company wants to use a 10% significance level. Ho: Ha: Decision Rule: Decision: Conclusion:

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4.
A fertilize company tests blends of fertilizer with hybrid seeds to
accurately determine yields. Consistency is desired, so the fertilizer
company wants to test the variability of yields. In years past, the farm
harvested an average of 88 bushels per acre, with a standard deviation of 6
bushels per acre.
acre plots the fertilize company collected data suggesting that the standard
deviation of the yield was 5 bushels per acre.
blend higher than the old?
This year, the farm altered the blend a bit. On 20 one-
Is the variance of the new
The company wants to use a 10% significance level.
Ho:
Ha:
Decision Rule:
Decision:
Conclusion:
Transcribed Image Text:4. A fertilize company tests blends of fertilizer with hybrid seeds to accurately determine yields. Consistency is desired, so the fertilizer company wants to test the variability of yields. In years past, the farm harvested an average of 88 bushels per acre, with a standard deviation of 6 bushels per acre. acre plots the fertilize company collected data suggesting that the standard deviation of the yield was 5 bushels per acre. blend higher than the old? This year, the farm altered the blend a bit. On 20 one- Is the variance of the new The company wants to use a 10% significance level. Ho: Ha: Decision Rule: Decision: Conclusion:
Expert Solution
Step 1

Chi-Square Test:

The hypothesis testing of the variability of the standard deviation is done by the chi-square test. The test is to compare the change in the variance of the population or of the sample with some treatment applied. The test statistic is calculated as the ratio of the standard deviation without treatment and with treatment. The critical value is calculated from the Chi-Square Table with alpha as the level of significance and the degree of freedom as N-1 (N is the size of the population/sample).

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