Part 2 (20pts) The casino goes ahead and purchases the 20 slot machines and want
ID: 3231850 • Letter: P
Question
Part 2 (20pts) The casino goes ahead and purchases the 20 slot machines and wants you to monitor their performance over the course of one month and make an inference about what the true-population mean is for how much the machines manufactured by IGT will return each month over its life So, you go ahead and place the 20 machines on the gaming floor and wait 1month. You then use software to determine for each machine how much it made (or lost) over that 1 month time period. You find that on average the 20 machines made $2,500 each over the one-month period, with a standard deviation across the 20 machines $100. 9. What is the general formula you should use for computing a confidence interval for u (2pt) Answer: 10. How many degrees of freedom do you have? (2pt) Answer: df 11. What is critical t for 95% confidence? (3pt) 13. What is the 95% confidence interval? (4pt Answer: Answer: 12. What is critical t for 85% confidence? (3pt) 14. What is the 85% confidence interval? (4pt) Answer: Answer: 15. How likely is it that in fact the true population mean is outside of your 85% confidence interval?(2pt) Answer:Explanation / Answer
Solution:-
Part 3)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 2450
Alternative hypothesis: 2450
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 22.36
DF = n - 1 = 20 - 1
D.F = 19
t = (x - ) / SE
t = 2.24
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 19 degrees of freedom is less than - 2.24 or greater than 2.24.
Thus, the P-value = 0.03724
Interpret results. Since the P-value (0.03724) is less than the significance level (0.05), we have to reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that new machines you have purchased are significantly different from old ones.