Afruit grower wants to test a new spray that a manufacturer claims will reduce the loss due to insect damage. to test the claim, the grower sprays 200 trees with the new spray and 200 other trees with the standard spray. the following data were recorded.
new spray standard spray
meanyieldpertree kg 109. 104
variance s2. 470. 375
(a) do the data provide sufficient evidence to conclude that the mean yield per tree treated with the new spray, μ1, exceeds that for trees treated with the standard spray, μ2? use α = 0.05. (round your answers to two decimal places.)
1-2. null and alternative hypotheses:
h0: (μ1 − μ2) = 0 versus ha: (μ1 − μ2) > 0h0: (μ1 − μ2) = 0 versus ha: (μ1 − μ2) ≠ 0 h0: (μ1 − μ2) ≠ 0 versus ha: (μ1 − μ2) = 0h0: (μ1 − μ2) < 0 versus ha: (μ1 − μ2) > 0h0: (μ1 − μ2) = 0 versus ha: (μ1 − μ2) < 0
3. test statistic: z =
4. rejection region: if the test is one-tailed, enter none for the unused region. z > z <
5. conclusion:
h0 is rejected. there is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. h0 is not rejected. there is insufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. h0 is rejected. there is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray. h0 is not rejected. there is sufficient evidence to conclude that the mean yield per tree treated with the new spray exceeds that for trees treated with the standard spray.
(b) construct a 95% confidence interval for the difference between the mean yields for the two sprays. (round your answers to two decimal places.)
kg to kg