In: Statistics and Probability
A recent article in Newsweek claims that the average weight of
women entering college as freshman is 135lbs. The staff at your
paper thinks this estimate is high. At a significance of 0.05, test
the claim that the mean weight is actually lower than 135lbs. What
is the null and alternate hypothesis? What is the Critical Value,
t?? What is the Test Statistic, T? Conclusion?
Then calculate the P-value. Does p-value agree support your
conclusion? Why?
149.9 |
141.1 |
123.5 |
154.3 |
127.9 |
110.2 |
123.5 |
154.3 |
134.5 |
116.8 |
125.7 |
127.9 |
108.0 |
152.1 |
114.6 |
138.9 |
123.5 |
119.0 |
125.7 |
138.9 |
119.0 |
123.5 |
138.9 |
112.4 |
130.1 |
143.3 |
116.8 |
136.7 |
121.3 |
125.7 |
141.1 |
132.3 |
114.6 |
121.3 |
92.6 |
Here claim is that mean weight is less than 135 lbs
As we know null hypothesis always have equality sign, so
vs
For the given data n=35, and (using excel)
The t-critical value for a left-tailed test, for a significance level of ?=0.05 is
tc?=?1.691
Graphically
Test statistics is
As t statistics is in rejection region we reject the null hypothesis
The P-Value is 0.002836, hence we reject the null hypothesis
So P value also support the conclusion