Question

Conduct each hypothesis test below using both the critical value/rejection region and p-value methods (separate and label each method) and showing each of the 5 steps explicitly. Do not round any table values. Round test statistics to the nearest hundredth, critical values to 3 decimal places, and p-values to 4 decimal values.

Many computer buyers have discovered that they can save a considerable amount by purchasing a personal computer from a mail-order company - an average of $900 by their estimates. In a test of this claim, a random sample of 19 customers who recently purchased a PC though a mail-order company were contacted and asked to estimate the amount that they had saved by purchasing by mail. The mean and standard deviation of these 19 estimates were $865 and $50 respectively. Is there sufficient evidence to indicate that the average savings differs from the $900 claimed by mail-order PC buyers at alpha = .05?

Answer 1

Solution :

= 900

=865

s =50

n = 19

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 900

H_a :    900

Test statistic = t

= ( - ) / s / n

= (865-900) / 50 / 19

= -3.05

P (Z <-3.05 ) =0.0069

P-value = 0.0069

= 0.05  

p=0.0069<0.05, it is concluded that the null hypothesis is rejected.

The significance level is α=0.05

The critical value for a two-tailed test is t_c​=2.101.

The rejection region for this two-tailed test is R = (t: |t| > 2.101)

Reject the null hypothesis .

There is enough evidence to claim that the population mean μ is different than 900, at the 0.05 significance level.