Question

Use a significance level of 0.05 to test the claim that the average life of cell phones equals 5 years. This is done after a study where the following statistical data are collected: n = 27, (x bar) ̅ = 4.6 years and s = 1.9 years.

a) Indicates Ho Ha,

b) draw the graph,

c) find the critical value,

d) find the t-statistic,

e) performs the hypothesis test to reject or fail to reject the null hypothesis.

f) Find the interval at 95% confidence

All work must appear to receive credit

Answer 1

The provided sample mean is 4.6 and the sample standard deviation is s = 1.9, and the sample size is n = 27

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 5

Ha: μ ≠ 5

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

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is t_c = 2.056

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that |t| = 1.094 < t_c = 2.056, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.284, and since p = 0.284 > 0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 5, at the 0.05 significance level.

The number of degrees of freedom are df = 27 - 1 = 26 , and the significance level is α=0.05.

Based on the provided information, the critical t-value for α=0.05 and df = 26 degrees of freedom is t_c = 2.056

The 95% confidence for the population mean μ is computed using the following expression

Therefore, based on the information provided, the 95 % confidence for the population mean μ is

CI = (3.848, 5.352)