54.20, 51.99, 46.96, 50.91, 51.05, 45.75, 46.80, 50.57, 52.45, 48.41, 53.39, 51.97, 50.37, 51.41 Research has shown that wing beat speed is normally distributed. So, we assume that our sample comes from a normal population with an unknown mean of u and an unknown standard deviation of a.. We would like to test whether the average wing beat speed of ruby-throated hummingbirds is greater than 50 beats per second. The null hypothesis is thus Ho:4=50. We will test this against the alternative H,. We want to test at the 6% level. Let x = the sample mean and s = the sample standard deviation. a) What should the alternative hypothesis, H, , be? O HaiH=6% O HaiH>50 O HaiH#50 O Haifl=50 O Hgifl<50 b) What is the formula for your test statistic? OT= X-6% x-50 OT= x-50 OT= V13 x-50 OT= x-50 OT=

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c) What value does your test statistic,T, take on with the sample data?
d) What type of probability distribution does your test statistic,T, have?
O t
O normal
O Chi-Squared
O binomial
O Cauchy
Transcribed Image Text:c) What value does your test statistic,T, take on with the sample data? d) What type of probability distribution does your test statistic,T, have? O t O normal O Chi-Squared O binomial O Cauchy
The ruby-throated hummingbird beats its wings very quicky. Using a high-speed camera, an ornithologist was able to measure the wing beating speed (in beats per second) of 14 randomly chosen ruby-throated
hummingbirds. The following are the measurements:
54.20, 51.99, 46.96, 50.91, 51.05, 45.75, 46.80, 50.57, 52.45, 48.41, 53.39, 51.97, 50.37, 51.41
Research has shown that wing beat speed is normally distributed. So, we assume that our sample comes from a normal population with an unknown mean of µ and an unknown standard deviation of o. . We
would like to test whether the average wing beat speed of ruby-throated hummingbirds is greater than 50 beats per second.
The null hypothesis is thus Ho:H=50 . We will test this against the alternative Ha .
We want to test at the 6% level.
Let x
the sample mean and s =
the sample standard deviation.
a) What should the alternative hypothesis, Ha , be?
O Hail=6%
Ο Ha μ50
О наи+50
Ο H μ=50
O HaiH<50
b) What is the formula for your test statistic?
OT =
x -6%
х-50
OT =
13
х-50
OT =
13
x-50
OT =
14
x-50
OT =
S
14
Transcribed Image Text:The ruby-throated hummingbird beats its wings very quicky. Using a high-speed camera, an ornithologist was able to measure the wing beating speed (in beats per second) of 14 randomly chosen ruby-throated hummingbirds. The following are the measurements: 54.20, 51.99, 46.96, 50.91, 51.05, 45.75, 46.80, 50.57, 52.45, 48.41, 53.39, 51.97, 50.37, 51.41 Research has shown that wing beat speed is normally distributed. So, we assume that our sample comes from a normal population with an unknown mean of µ and an unknown standard deviation of o. . We would like to test whether the average wing beat speed of ruby-throated hummingbirds is greater than 50 beats per second. The null hypothesis is thus Ho:H=50 . We will test this against the alternative Ha . We want to test at the 6% level. Let x the sample mean and s = the sample standard deviation. a) What should the alternative hypothesis, Ha , be? O Hail=6% Ο Ha μ50 О наи+50 Ο H μ=50 O HaiH<50 b) What is the formula for your test statistic? OT = x -6% х-50 OT = 13 х-50 OT = 13 x-50 OT = 14 x-50 OT = S 14
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