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
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.