The water pollution readings at State Park Beach seem to be lower than that of t
ID: 3223153 • Letter: T
Question
The water pollution readings at State Park Beach seem to be lower than that of the prior year. A sample of 12 readings (measured in coliform/100 mL) was randomly selected from the records of this year's daily readings. Does this sample provide sufficient evidence to conclude that the mean of this year's pollution readings is significantly lower than last year's mean of 3.8 at the 0.05 level? Assume that all such readings have a normal distribution.
(ii) Find t. (Give your answer correct to two decimal places.)
Explanation / Answer
Sample mean=47.5/12=3.96
Sample variance =S=.654
Which is greater than prior year mean 3.8...
Now t statistics for unknown variance
, we know that if the data are normally distributed, then:
T=(X bar) - /S*n
follows a t-distribution with n1 degrees of freedom. Therefore, it seems reasonable to use the test statistic:
T=X¯0 / S*n
for testing the null hypothesis H0:=0H0:=0 against any of the possible alternative hypotheses HA:0HA:0, HA:<0HA:<0, and HA:>0HA:>0. For the example in hand, the value of the test statistic is:
t= 3.8 - 3.96 /12*.654 = -.07
The critical region approach tells us to reject the null hypothesis at the = 0.05 level if t t0.05,11 = 2. 201 or if t t0.05,11 = 2.201
Here t =-.07< 2.201
So we reject null hypothesis at 5% level of significance.
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