4) A high school biology student wishes to test the hypothesis
that hummingbird feeders can affect the mean mass of ruby-throated
hummingbirds in the area surrounding the feeder. She captures and
weighs several of the hummingbirds near a science museum where
several feeders are located. She obtains the following masses in
grams: 4.4, 3.9, 4.5, 4.3, 4.1, 3.8, 3.8, 4.1, 3.9, 3.8, 3.2, 4.3
The student's hypotheses are: H0: µ= 3.65 g Ha: µ> 3.65 g Use
technology to calculate the P-value, then determine whether the
data provide sufficient evidence to conclude that the mean mass of
the birds in the area surrounding the feeder is greater than the
mean mass of the general population. Test at the 5% significance
level and assume that the population standard deviation is 0.35 g.
Also, assess the strength of the evidence against the null
hypothesis.
4)
A) P = 0.0002; since P <0.05, reject the null hypothesis. At the
5% significance level, the data do provide sufficient evidence to
conclude that the mean mass of the birds in the area surrounding
the feeder is greater than the mean mass of the general population.
The evidence against the null hypothesis is very strong.
B) P = 0.0004; since P <0.05, do not reject the null hypothesis. At the 5% significance level, the data do not provide sufficient evidence to conclude that the mean mass of the birds in the area surrounding the feeder is greater than the mean mass of the general population. The evidence against the null hypothesis is weak or none.
How would i do this with the TI-84?
H0: µ= 3.65
Ha: µ> 3.65
calculate sample mean for the given data using calculator we get
sample mean ( ) = 4.0083
population standard deviation ( ) = 0.35
sample size ( n ) = 12 , = 5% = 0.05
now use following steps by using TI-84 calculator
1) Press button STAT
2) then navigate to TESTS using RIGHT ARROW KEY
3) then press button 1 for Z test
4) now insert data as follows
5) now navigate to button CALCULATE and press ENTER
we get output as follows
here P-Value = 1.9541837E-4 = 0.00019 = 0.0002
P-Value = 0.0002
A) P = 0.0002; since P <0.05, reject the null hypothesis. At the 5% significance level, the data do provide sufficient evidence to conclude that the mean mass of the birds in the area surrounding the feeder is greater than the mean mass of the general population. The evidence against the null hypothesis is very strong.
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