Answer to Question #100112 in Statistics and Probability for sofia

Question

A mobile phone manufacturer has designed a new model of mobile phone and has invited 40 customers to take part in a trial session. After the trial session, the customers are invited to state their comments on the phone as well as to indicate how much they are willing to pay for this mobile phone. The results indicate that on the average a customer is willing to spend $5500 for it. Assume that the population standard deviation is the same as the old model, which is $1150. The director of the manufacturer suggests selling this new mobile phone at $6000. Test, at 3% significant level, whether the result collected from the trial session indicates the population mean amount a customer is willing to pay for this mobile phone is lower than the director’s suggestion.

Answer 1

The provided sample mean is $\bar{X}$ and the known population standard deviation is $\sigma=1150,$ and the sample size is $n=40.$

The following null and alternative hypotheses need to be tested:

$H_0:\mu=6000$

$H_1:\mu<6000$

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 $\alpha=0.03,$ and the critical value for a left-tailed test is $z_c=-1.88$

The rejection region for this left-tailed test is $R=\{z:z<-1.88\}$

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}={5500-6000\over 1150/\sqrt{40}}=-2.75$$

Since $z=-2.75<-1.88=z_c,$  it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is $p=0.003.$ Since $0.003<0.03,$ it is concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis $H_0$ is rejected. Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 6000, at the 0.03 significance level.

Therefore, there is enough evidence to claim that the population mean amount a customer is willing to pay for this mobile phone is lower than the director’s suggestion, , at the 0.03 significance level.