The Crop, Soil, and Science Department states the average final exam score for S

Question

The Crop, Soil, and Science Department states the average final exam score for Soybean Production CSES 322 is 92. You suspect it is lower. From previous years, the population standard deviation is assumed to be 5.7. You take a random sample of 23 students and obtain the following exam scores:

S = {80,95,95,75,82,54,90,98,83,91,96,92,97,81,64,82,90,85,73,77,93,95,76}.

a) Define the random variable and test an appropriate set of hypotheses to address the question of interest by calculating a p-value and interpreting its numerical value in the context of the problem. Make your conclusions at the 5% level.

b) If your decision in part (a) was an mistake, what type of error did you commit?

c) What is the power of the test when the alternative a = 83?

Explanation / Answer

a) Let x be the sample mean. Assume that the samples are taken from the normal population. Then x is the random variable which follows normal distribution with mean and variance ².

We are testing H0 : =93 Vs H1: <92

If the null hypothesis is true, then the test statistic (x-93)n /     follows N(0,1).

Now x = (80.95+95.5+...+95.6)/23 = 84.52, n=23 =5.7,

So observed value of test statistic is = (84.52-93)23 / 5.7 = -7.135

p-value= P(Z<-7.135)=0<0.05 ,the significance level

Therefore at 5% significance level we reject the null hypothesis and conclude that population mean is less than 93 based on the sample we got.

b) If the decision in part (a) is a mistake then we have commited Type1 error. Type1 error is rejecting H0 when H0 is true.

c) Power of the test is the probability of rejecting H0 when H1 is true.

    So power = P(x<(93-1.645*(/n)) I =83), Since 5%lower critical value is -1.645

                   = P(x<(93-1.645*(5.7/23)) I =83)

                    = P(x<91.04 I =83)

                    = P((x-83)n/<(91.04-83)n/)

                     =P((x-83)23/5.7<(91.04-83)23 /5.7)

                     =P(Z<6.765)

                     1

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