Answer to Question #115474 in Statistics and Probability for Windia nababan

Question

Suppose that an airline company claims that the average weight of checked baggage is less than 15 pounds. To support the claim, the airline company conducts a random sample of 150 passengers and finds that the average weight of checked baggage is 14.2 pounds, with a standard deviation of 6.5 pounds. Do these data support the airline company's claim at the α = 0.05 level of significance? Use s = 6.5 as an estimate for since n 30.

Answer 1

The provided sample mean is $\bar{X}=14.2$  and the known population standard deviation is $\sigma=6.5,$ and the sample size is $n=150>30.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu\geq15$

$H_1:\mu<15$

This corresponds to a left-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 $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.645.$

The rejection region for this left-tailed test is $R=\{z:z<-1.645\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu\over \sigma/\sqrt{n}}={14.2-15\over6.5/\sqrt{150}}\approx-1.507378$$

Since $z=-1.507378>-1.645=z_c,$ it is observed that  it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population $\mu$  is less than 15, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.0659,$ and since $p=0.0659\geq0.05=\alpha,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population $\mu$ is less than 15, at the 0.05 significance level.