1stats help please 2?

Question

Bill claims that the mean weight of the population of bears living in Yellowstone National Park is 210 pounds. A sample of 64 bears from Yellowstone has a mean weight of 192.9 pounds with s=123.2. Does the claim hold at the 0.05 significance level?

Answer 1

the claim is plausible.

Hypothesis Test for mean:

Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis

H0: μ ≤ Δ or

H0: μ ≥ Δ or

H0: μ = Δ

Find the test statistic z = (xbar - Δ ) / (sx / √ (n))

where xbar is the sample average

sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.

n is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z

H1: μ < Δ; p-value is the area to the left of z

H1: μ ≠ Δ; p-value is the area in the tails greater than |z|

If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.

If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible.  Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ ≥ 210 vs. H1: μ < 210

The test statistic is:

z = ( 192.2 - 210 ) / ( 123.2 / √ ( 64 ))

z = -1.155844

The p-value = P( Z < z )

= P( Z < -1.155844 )

=  0.1238725

Since the p-value is greater than the significance level we fail to reject the null hypothesis and conclude μ ≥ 210 is plausible.

Answer 2

Assuming a normal distribution of your data...

You would look for your Z score to get your 95% (1-.05) confidence level (in this case it would be appx 1.64). Your formula would be: mean + 1.64*123.2. This gives you 95% of the bear weights at Yellow stone (1-.05). In this case, the value is ~394 lbs, so the assumption holds at the .05 significance level.