Test Sample Vs. The Benchmark With 98 % Confidence

As noted in the first table of the clerical test, the mean score in the results for the clerical test are very similar in the two tests made by the company, in Time one the mean was 31.61

Standard deviation is important in comparing two different sets of data that has the same mean score. One standard deviation may be small (1.85), where the other standard deviation score could be quite large (10)(Rumsey,

The numbers given in the case were as follows: assumed population standard deviation is equal to .21, sample size is equal to 30 and the test value of the mean was 12. They also stated the two hypotheses to be tested: the null hypothesis that the population is equal to 12 and the alternative hypothesis that the mean is not equal to 12.

(21) You took a sample of size 21 from a normal distribution with a known standard deviation, . In order to find a 90% confidence interval for the mean, You need to find.

Angus Reid used a margin of error correctly as you stated, “The report from the Angus Reid Institute clearly stated that the margin of error was used for "comparison purposes only."”

Solution: since the sample size 81>30 therefore we use Z test and the test statistics will be

A sample of 54 day-shift workers showed that the mean number of units produced was 345, with a standard deviation of 21. A sample of 60 afternoon-shift workers showed that the mean number of units produced was 351, with a standard deviation of 28 units. At the .05 significance level, is the number of units produced on the afternoon shift larger?

c) Is there a high probability that the mean and standard deviation of your sample are consistent

13. For the scores (3.0, 3.4, 2.6, 3.3, 3.5, 3.2): N = 6; Sum = 19; Mean = Sum/N = 19/6 = 3.17; Median = (2.6+3.3)/2 = 2.95; Sum of Squared Deviations = (3-3.17)2+(3.4-3.17)2+(2.6-3.17)2+(3.3-3.17)2+(3.5-3.17)2+(3.2-3.17)2 = 0.533; SD2 = SS/N = 0.533/6 = 0.089; SD = 0.089 = 0.298.

Suppose the significance level for a hypothesis test is α = 0.05. If the p-value is 0.049, the decision is to

The interval contained 13.4% III. A 99% confidence interval estimate contained 13.4% A. I and II B. I and III C. II and III D. I, II, and III E. None of the above gives the complete set of true responses. 14. The margin of error in a confidence interval estimate using z-scores covers which of following? I. Random sampling errors II.

Construct a 99% confidence interval for the population standard deviation, . A) (0.8, 2.1) B) (1.0, 2.6) C) (0.9, 2.5) D) (1.0, 2.4) Answer: C B) 2.35-2.55 C) 2.35-2.54 D) 2.35-2.75 18) 19) 3 20) In a survey of 2480 golfers, 15% said they were left-handed. The survey 's margin of error was 3%. Construct a confidence interval for the proportion of left-handed golfers.

Both data sets contain outliers, as clearly shown in the box and whisker plots. The control group contains two outliers, both of which are 70. The experimental group contains four outliers: 0, 10, 40 and 70. In all calculations, the outliers have been omitted in order to ensure more accurate calculations. The median test score/center of the data for both the control group and experimental group is around 90 percent. The mean or average test score for the experimental group is 93.5 and for the control group, the average test score is an 89.3. The standard deviation of the group that chewed gum is 6.895 (about 7) and the group that did not chew gum had a standard deviation of 7.164 (about 7.2.) This shows that the experimental group has less variance in comparison to the control group. Without doing an actual hypothesis test, just based off of the graphs and raw data, we can guess that our original hypothesis (those who chew gum will have better test scores than those who do not) might be correct and the conclusion will be that we fail to reject the null hypothesis. Based on the dotplot and histogram, the assumption of normality is plausible for the control group but not for the experimental

The p-value for the hypothesis testing was lower than 0.05, thus, the null hypothesis is rejected and accepting the alternative hypothesis.

We are looking at a very large group of tests. Forty five tests were selected from this group. The mean of the sampled tests is 58 and the standard deviation is 13. Find the 95% confidence interval for the mean grade for the whole class.