1.To test the effectiveness of a new spray for controlling rust mites, we would
ID: 3354030 • Letter: 1
Question
1.To test the effectiveness of a new spray for controlling rust mites, we would like to compare the average yield for treated groves with the average yield displayed for untreated groves in previous years. A random sample of 30 one-acre groves is chosen and sprayed according to a recommended schedule. The average yield for the sample of 30 groves was 830 boxes, with a standard deviation of 91. Yields from groves in the same area without rust mite maintenance spraying has averaged 760 boxes over previous years. Do these data present sufficient evidence to indicate that the mean yield for groves sprayed with the new preparation is higher than 760 boxes, the average over previous years without spraying? Use = 0.05.
H0:
Ha:
Test Statistic:
p-value:
Conclusion:
Explanation / Answer
Solution:-
= 830, = 91, n = 30
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 760
Alternative hypothesis: > 760
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 16.6143
DF = n - 1
D.F = 29
t = (x - ) / SE
t = - 4.21
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 4.21.
Thus the P-value in this analysis is 0.00011.
Interpret results. Since the P-value (0.00011) is less than the significance level (0.05), we have to reject the null hypothesis.
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