Data shows the times for carrying out a blood test at Rivervalley Labs
Using Excel Plot a histogram of the data. What type of distribution does the data appear to follow?
Construct and interpret the 95% confidence interval for the population mean time for carrying out a blood test. Assume that the population standard deviation is unknown.
Times for blood tests (minutes)
2.5 |
4.9 |
3.6 |
4.3 |
1.9 |
3.4 |
3.2 |
4.0 |
3.1 |
3.6 |
3.9 |
4.4 |
3.9 |
3.1 |
2.7 |
3.5 |
3.6 |
3.7 |
4.9 |
3.6 |
3.8 |
3.6 |
2.5 |
4.0 |
3.6 |
4.0 |
3.2 |
3.5 |
2.6 |
3.7 |
4.9 |
4.0 |
4.2 |
2.6 |
4.4 |
3.5 |
4.8 |
5.0 |
3.5 |
3.2 |
3.6 |
2.6 |
3.1 |
5.0 |
2.0 |
4.3 |
5.0 |
3.9 |
3.5 |
5.3 |
3.5 |
4.0 |
4.1 |
3.7 |
4.5 |
3.7 |
2.8 |
3.4 |
2.8 |
2.5 |
2.9 |
2.4 |
5.0 |
3.0 |
4.1 |
3.6 |
3.8 |
3.2 |
4.3 |
3.2 |
3.8 |
4.1 |
1.7 |
4.3 |
4.7 |
3.2 |
3.1 |
3.6 |
2.1 |
3.4 |
2.1 |
3.4 |
3.3 |
3.8 |
2.6 |
2.5 |
3.0 |
4.3 |
3.8 |
3.4 |
1.9 |
3.0 |
4.5 |
4.0 |
4.6 |
4.6 |
3.9 |
1.7 |
3.9 |
4.3 |
Upper Limit | Frequency |
0.5 | 0 |
1 | 0 |
1.5 | 0 |
2 | 5 |
2.5 | 7 |
3 | 11 |
3.5 | 22 |
4 | 29 |
4.5 | 14 |
5 | 11 |
5.5 | 1 |
6 | 0 |
6.5 | 0 |
7 | 0 |
The nature of the histogram suggests that the distribution is bell-shaped and approximately symmetrical. So, a normal distribution assumption may be valid for this distribution.
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Assume that the data is pasted on Column A of an Excel worksheet.
The sample mean = AVERAGE(A:A) = 3.578
The sample SD = STDEV.S(A:A) = 0.812
Z = NORMSINV(1 - (0.05/2)) = 1.96
Confidence Interval: Sample mean +/- Z*sample SD/sqrt(N)
= 3.578 +/- 1.96*0.812/sqrt(100) = [3.42,
3.74]
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