Question

Utilizing a random sample of 100, the Richardson marketing research firm would like to test whether it is true that the mean customer satisfaction rating for the new virtual reality system exceeds 50 (benchmark mu). After collecting the data, you, the analyst, find that the sample mean is 50.99 and the sample standard deviation (s) is 3.360.

(a) Build H0 (null hypothesis) and Ha (alternative hypothesis) for testing the claim that the mean customer satisfaction rating is above 50. Null, H0: µ ≤50        Alt, H1: µ >50

(b) Calculate the t statistic. 2.946

(c) Using the p-value approach to perform the t test on H0 versus Ha at each of the alpha level 0.10, 0.05, 0.01, and 0.001.

=.002

(d) How much evident is there that the mean customer satisfaction rating will be above 50?

a=.10,.05 and .01 accepting the claim that the mean rating is above 50.

a=.001 reject the null hypothesis that the satisfaction is 50 or less

evidence in the 95% confidence range

(e) What recommendations would you offer to the manager, based on the hypothesis testing results? New VR system exceeds 50 at .10 .05 and .01. We would prefer these levels of significance to the claim.

test statistic
t	(sample mean-mu)/(s/sqrt(n))				mu is the benchmark
critical value
t	t.inv
p value	cumulative true
t	t.dist

Answer 1

Let X : Customer Satisfaction rating for new virtual reality system.

To test:

H0: mu<= 50 vs H1: mu> 50

Since at 0.10, 0.05, 0.01 level of significance p value < 0.1/0.05/0.01. Thus H0 is rejected at all this 3 los value.

e. Thus customer satisfaction rating exceeds 50 for a new virtual reality system.

I have given my best to solve your problem. Please like the answer if you are satisfied with it. ?