Question #215644

A melting point test ofinwant to test the claim that the temperature is different fromx D154:2 F. Assume that the melting point is normally distributed withn D10 samples of a binder used in manufacturing a rocket propellant resulted155 F: Test the hypothesis atD1:5 F: Suppose youD0:05:

1
Expert's answer
2021-07-12T03:36:24-0400

The following null and alternative hypotheses need to be tested:

H0:μ=155H_0:\mu=155

H1:μ155H_1:\mu\not=155

​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 isR={z:z>1.96}.R=\{z:|z|>1.96\}.

The z-statistic is computed as follows:



z=xˉμσ/n=154.21551.5/10z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{154.2-155}{1.5/\sqrt{10}}1.686548\approx-1.686548

Since it is observed that z=1.686548<1.96=zc,|z|=1.686548<1.96=z_c, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value is p=2P(Z<1.686548)=0.09169,p=2P(Z<-1.686548)=0.09169, and since p=0.09169>0.05=α,p=0.09169>0.05=\alpha, it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ\mu

is different than 155, at the α=0.05\alpha=0.05 significance level.


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