statistics- hypothesys

For your fourth journal entry, consider again the variable DP05. Last week you constructed a 95% confidence interval for the number of months that had no days with 0.5 inches or more of precipitation.  A resident thinks that 15% of the months have no days with 0.5 inches or more of precipitation. Test this claim at a 0.05 significance level. How do your results compare to your 95% confidence interval from the Module Three journal? 

Hypothesis test results: Outcomes in : DP05 Success : 0 p : Proportion of successes H0 : p = 0.15 HA : p ≠ 0.15

Variable	Count	Total	Sample Prop.	Std. Err.	Z-Stat	P-value
DP05	36	347	0.1037464	0.019168598	-2.4129882	0.0158

Now consider the variable EMNT. This is the extreme low temperature (extreme minimum temperature) for the month, also reported in tenths of a degree Celsius. For example, a value of –50 in EMNT means that the extreme low for the month was –5 degrees Celsius, or 23 degrees Fahrenheit.  A resident thinks that the extreme low temperature during a randomly chosen month is about –5 degrees Celsius. Test this claim at a 0.01 significance level. Use both the classical method and the p-value method. Does this analysis make sense? 

Hypothesis test results: μ : Mean of variable H0 : μ = 50 HA : μ ≠ 50

Variable	Sample Mean	Std. Err.	DF	T-Stat	P-value
EMNT	-55.048991	6.181248	346	-16.994787	<0.0001