Question #113659
We are interested to know whether test animals respond to temperature by moving
toward a favoured temperature. Animals were placed in a temperature gradient marked
in units from zero (the starting point) in the middle of the gradient to positive at the
warm end and negative at the cold end. The mean position along the gradient was found
to be -1.352 units, the standard deviation was 12.267 units and n=500 individuals. Is
there a significant tendency to aggregate at either the cold or warm end of the gradient?
Start by setting up a null hypothesis in terms of the mean position along the gradient =
0. (This question is graded out of 12 marks)
1
Expert's answer
2020-05-04T19:29:10-0400

The Central Limit Theorem

Let X1,X2,...,XnX_1,X_2,...,X_n be a random sample from a distribution with mean μ\mu and variance σ2.\sigma^2. Then if nn is sufficiently large, Xˉ\bar{X} has approximately a normal distribution with μXˉ=μ\mu_{\bar{X}}=\mu and σXˉ2=σ2/n.\sigma_{\bar{X}}^2=\sigma^2/n.

If n>30,n>30, the Central Limit Theorem can be used.

The provided sample mean is Xˉ=1.352\bar{X}=-1.352 and the known population standard deviation is σ=12.267,\sigma=12.267, and the sample size is n=500.n=500.

The following null and alternative hypotheses need to be tested:

H0:μ=0H_0:\mu=0

H1:μ0H_1: \mu\not=0

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha=0.05, and the critical value for a two-tailed test is zc=1.96.z_c=1.96.

The rejection region for this two-tailed test is R={z:z>1.96}R=\{z:|z|>1.96\}

The z-statistic is computed as follows:


z=Xˉμ0σ/n=1.352012.267/5002.464469z={\bar{X}-\mu_0 \over \sigma/\sqrt{n}}={-1.352-0 \over 12.267/\sqrt{500}}\approx-2.464469

Since it is observed that z=2.464469>1.96=zc,|z|=2.464469>1.96=z_c, it is then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean μ\mu  is different than 0, at the 0.05 significance level.

Using the P-value approach: The p-value is p=0.0137,p=0.0137, and since p=0.0137<0.05=α,p=0.0137<0.05=\alpha, it is concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population mean μ\mu  is different than 0, at the 0.05 significance level.

There a significant enough evidence to claim that animals respond to temperature by moving toward a favoured temperature, at the 0.05 significance level.



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