Question

A set of numbers such as those below is meaningless without some background information. The INDIVIDUALS here are Newcomb's 66 repetitions of his experiment. We need to know exactly WHAT VARIABLE he measured, and in WHAT UNITS. Newcomb measured the time in seconds that a light signal took to pass from his laboratory on the Potomac River to mirror at the base the Washington Monument an back, a total distance of about 7400 meters. Just as you can compute the speed of a car from the time required to drive a mile, Newcomb could compute the speed of light from the passage of time. Newcomb's first measurement of the passage of time of light was 0.000024828 second or 24828 nanoseconds. The entries above record only the deviation from 24800, so the first entry, 28, is short for 24828 nanoseconds. Negative entries are numbers less than 24800.

28	22	36	26	28	28
26	24	32	30	27	24
33	21	36	32	31	25
24	25	28	36	27	32
34	30	25	26	26	25
-44	23	21	30	33	29
27	29	28	22	26	27
16	31	29	36	32	28
40	19	37	23	32	29
-2	24	25	27	24	16
29	20	28	27	39	23

Use the data set above to make a histogram and analyze it, using key terms

Present a 5 number summary and a modified box plot and are there any outliers?

Report the mean and standard deviation (do not discard outliers), the mean was important in this experiment. Calculate the 95% confidence interval for the true mean. Explain what this means.

Compare these (5 number summary and mean/standard deviation). Are the mean and standard deviation valid for this set of data? Justify your answer.

Some of the above (and what follows below) makes no sense if the data is not approximately normal. Explain what this means. Is this data close to normal distributed? Justify your answer. Regardless of your conclusion, for the next part assume the data is approximately normal.

The data listed in the order it was recorded (down first, then across).

Do a time plot. Analyze this plot, paying close attention to new information gained beyond what we did above.

Cut the data in half (first three columns vs. last three columns) and do a back to back stem plot. Analyze this. Does this further amplify what the time plot showed?

Calculate the mean of the second half of the data.

Using the mean and standard deviation of the whole data set (found above) as the population mean and standard deviation, test the significance that the mean of the second half is different than the mean of the total using α=.05 .

Make sure to clearly identify the null and alternative hypothesis.

Explain what this test is attempting to show.

Report the p-value for the test and explain what that means.

Accept or reject the null hypothesis, and justify your answer (based on the p-value).

Answer 1

most of the frequency distributed within 20 to 40 class interval.there is one outlier. total 24 frequency distributed wihin 25 to 30 class interval.

yes, there is one outlier( star symbol)

mean = 27.31034

standard deviation = 6.365876

confidence interval = mean z* SE

MEAN = 27.31034

Z = 1.96 at 95 % confidence level

SE = standard error = standard deviation / square root of sample size(66)

= 27.31034 1.96 * 0.7836

= 27.310341.5358

means true mean vary from 25.7745 to 28.846

yes mean and the standard deviation is valid in this data set because standard error is very less and confidence interval does not vary so much.

means above confidence interval does not make sense if data are not normally distributed

mean = 27.31034

median =27

mode = 27

skewness = -1.51311

yes data are close to normal distribution because mean , median and mode value are approximately same but slightly leftward distributed because skewness is -1.51311

after cut half data

after the cut, there is no outlier , most of the data distributed within 25 to 30 class interval.

mean = 27.96296

standard deviation = 4.316047

null hypothesis = there is a no significant difference between the mean of the second half and total

alternative hypothesis = there is a difference between the mean of the second half and total

z stat use to prove this hypothesis

z =

X1 bar = 27.31034

X2 bar =27.96296

S1 = 6.365876

S2 = 4.316047

N1 =N2=66

Z= -0.6894

The P-Value is .245286. at one tail test

so, accept the null hypothesis because p value is > 0.05.