An article in the San Jose Mercury Mews stated that students
Solution
(77) Let mu be the population mean
The test hypothesis:
Ho: mu=4.5 (i.e. null hypothesis)
Ha: mu > 4.5 (i.e. alternative hypothesis)
The test statistic is
Z=(xbar-mu)/(s/vn)
=(5.1-4.5)/(1.2/sqrt(49))
=3.5
It is a right-tailed test.
Given a=0.01, the critical value is Z(0.01) = 2.33 (from standard normal table)
The rejection region is if Z>2.33, we reject the null hypothesis.
Since Z=3.5 is larger than 2.33, we reject the null hypothesis.
So we can conclude that the data support my claim
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(80) Ho: p=0.6
Ha: p not equal to 0.6
The test statisitc is
Z=(phat-p)/sqrt(p*(1-p)/n)
=(34/64-0.6)/sqrt(0.6*0.4/64)
=-1.12
It is a two-tailed test.
Assume that the significant level a=0.05
The critical values are Z(0.025) = -1.96 or 1.96 (from standard normal table)
The rejection regions are if Z<-1.96 or Z>1.96, we reject the null hypothesis.
Since Z=-1.12 is between -1.96 and 1.96, we reject the null hypothesis.
So we can not conclude that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched
