Question

1. If X is non-normal and n<30, then the sampling distribution of standardized X-bar is:

1. t (n-1)
2. approximately Z
3. unknown
4. binomial

2. A statistician wants to estimate the mean loss suffered by Delta pilots in the labor negotiations to lower Delta salaries to within $2500 with 95% confidence. From a first small survey, the standard deviation of the loss is estimated at $10,000. What size sample should the statistician select?

1. 1206
2. 44
3. 7
4. None of the above

3. The difference in the observed neurological disease rate, xbar, for a sample of veterans who served in Iraq is not “statistically significantly different” (alpha = .05), from the overall population disease rate for all U.S. veterans, mu₀. This means

1. xbar = mu₀
2. The disease rates xbar and mu₀ are not equal, but the size of the difference is not practically important, not big enough to matter
3. H₀: mu for Iraq vets = mu₀   was not rejected
4. None of the above

4. Suppose that the population of SAT scores is normally distributed with a mean of 1000 and a standard deviation of 100. To determine the effect of a course to prepare for the SAT, a random sample of 25 students who have taken the course is selected. The sample mean SAT is 1050. Do these data provide sufficient evidence at the 1% significance level to infer that students who take the course perform better on the SAT on average? Assume that the population standard deviation of scores did not change.

1. No, the p-value is greater than .01
2. Yes, the p-value is greater than .01
3. Yes, the p-value is less than .01
4. None of the above

Answer 1

1. c) . unknown

Because, t test require normality assumption with small size and z-test require normality assumption with size > 30.

Here, we standardize the scores which means that we are just changing their scale so that they will have mean 0 and standard deviation 1 but it doesnot change how the scores are distributed in their range. Hence, distribution remains non normal and unknown and when we dont know the distribution we cannot apply a test.

please rate

- for mean (I is x I zde o for a = 0. for one failed, zal = 1.645, 6= $ 10000 & limits = $2500 . 1:645 * $ 1900 zin of n = (6:6)² = 43.56 $2500 or no ~44 as 6 n is 44 2 Ente I I 3 3 Lance the difference between mean from sample t and mean from poppy n is not statistically significant'. we can say they are equal or @ T = 1o as @ xbar = & mus. 3 EXP. 3 U-1000, 1=25, 6= 100, =1050, x=0.01 Ans DRGE 2 = 2.50 nting t= nu = so in 10/55 here, df = 25-1 = 24 (it is one tailed test) 2.797 tuintal = and fralue> 0.01 (from t-disth table). o we reject Ho. or we can say @ No, the pvalue is greater than 0.01