(1 pt) A random sample of 100 observations from a population with standard devia
ID: 3361148 • Letter: #
Question
(1 pt) A random sample of 100 observations from a population with standard deviation 22.6881507785873 yielded a sample mean of 93.8. (a) Given that the null hypothesis is -90 and the alternative hypothesis is > 90 using a = 05, find the following: (i) critical z score (ii) test statistic (b) Given that the null hypothesis is = 90 and the alternative hypothesis is 90 using = 05 find the following (i) the positive critical z score (ii) the negative critical score (ii) test statistic The conclusion from part (a) is: O A. There is insufficient evidence to reject the null hypothesis O B. Reject the null hypothesis O C. None of the above The conclusion from part (b) is: A. There is insufficient evidence to reject the null hypothesis O B. Reject the null hypothesis 0 C. None of the aboveExplanation / Answer
(i) critical z score for 1 tailed test at 0.05 level of significance z=1.6449
(ii) std error =std deviation/n)1/2 =22.68815/(100)1/2 =2.268815
therefore tes statistic z=(X-mean)/std error =(93.8-90)/2.268815=1.6749
b)
(i)the positive critical z score =1.9600
(ii)the negative critical z score =-1.9600
(iii) test statistic z ==(93.8-90)/2.268815=1.6749
conclusion from part (a) is option B) reject the null hypothesis
conclusion from part (b) is option A) there is insufficient evidence .............
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