Question

In a study of cereal leaf beetle damage to oats, researchers measured the number of beetle larvae per stem in small plots of oats after applying (or not applying) the pesticide Malathion . Researchers applied Malathion to a random sample of 5 plots, and did not apply the pesticide to an independent sample of 12 plots. A noted scientist claims that Malathion will not make any difference in the mean number of larvae per stem. Test her claim at the .05 level of significance. State H0 and H1. a. H0: Mean Malathion -MeanNoMalathion <= 0 H1: Mean Malathion-Mean Malathion > 0 b. H0: MeanMalathion -MeanNoMalathion >= 0 H1: MeanMalathion-MeanNoMalathion < 0 c. H0: MeanMalathion = MeanNoMalathion H1: MeanMalathion ≠ MeanNoMalathion d. H0: MeanMalathion-MeanNoMalathion = 0 H1: MeanMalathion-MeanNoMalathion ≠ 0

Answer 1

Answer:

At 0.05 level of significance, there is no difference in the mean number of larvae per stem. This supports the scientist's claim.

(d) H0: MeanMalathion - MeanNoMalathion equals 0

H1: MeanMalathion - MeanNoMalathion not equals 0

Step-by-step explanation:

Test statistic (t) = (mean 1 - mean 2) ÷ sqrt[pooled variance (1/n1 + 1/n2)]

Let the difference between the two means be x and the pooled variance be y

n1 = 5, n2 = 12

t = x ÷ sqrt[y(1/5 + 1/12)] = x ÷ sqrt(0.283y) = x ÷ 0.532√y = 1.88x/√y

Assuming the ratio of x to √y is 0.5

t = 1.88×0.5 = 0.94

n1 + n2 = 5 + 12 = 17

degree of freedom = n1 + n2 - 2 = 17 - 2 = 15

significance level = 0.05 = 5%

critical value corresponding to 15 degrees of freedom and 5% confidence interval is 2.131

The test is a two-tailed test because the alternate hypothesis is expressed using not equal to.

The region of no rejection of the null hypothesis lies between -2.131 and 2.131

Conclusion

Fail to reject the null hypothesis because the test statistic 0.94 falls within the region bounded by the critical values.

The scientist's claim is right.

A null hypothesis is a statement from a population parameter which is either rejected or accepted (fail to reject) upon testing. It is expressed using the equality sign.

An alternate hypothesis is also a statement from a population parameter which negates the null hypothesis and is accepted if the null hypothesis is rejected. It is expressed using any of the inequality signs.