Question

1)In a study designed to assess the side effects of two drugs, 50 animals were given drug A and 50 animals were given drug B. Of the 50 receiving drug A,10 showed undesirable side effects, while 8 of those receiving drug B reacted similarly. If there is no difference between the population proportions of animals showing undesirable side effects when taking drug A or drug B, the best estimate for the common proportion is?

2)A significance test was performed to test the null hypothesis H0 : μ = 3 versus

Ha : μ ̸= 3. The test statistic is z = −1.23. The P -value for this test is approximately ?

3)Suppose that cellulose content in a variety of alfalfa is normally distributed with un- known mean μ and standard deviation σ = 8mg/g. A sample of 16 plots of the hay yielded mean cellulose content x = 145mg/g. A 95% confidence interval for the mean cellulose content is?

Answer 1

Solution1:

best estimate for the common proportion pbar=x1+x2/n1+n2=10+8/50+50=18/100=0.18

Soluton2:

p value in excel is

2*left tail

left tail probability =NORMSDIST(-1.23)

=0.109348552

=2*0.109348552

=0.218697105

p=0.2187

Solution3:

z ccrit for 95%=1.96

95% confidence interval for the mean cellulose content i

xbar-z*sigma/sqrt(n),xbar+z*sigma/sqrt(n)

145-1.96*8/sqrt(16),145+1.96*8/sqrt(16)

141.08, 148.92

95% confidence interval for the mean cellulose content lies in between 141.08 and 148.92