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the following measurements were recorded for the drying time, in hours, of a certain brand of latex paint. 3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6 3.7 2.8 4.4 4.0 5.2 3.0 4.8 assume that the measurements are a simple random sample. (a) what is the sample size for the above sample? (b) calculate the sample mean for these data. (c) calculate the sample median. (d) plot the data by way of a dot plot. (e) compute the 20% trimmed mean for the above data set. (f) is the sample mean for these data more or less descriptive as a center of location than the trimmed mean?

2 Answers

1 vote

Answer:a) The sample size for the given data set is 15. This can be determined by counting the number of measurements recorded.

(b) To calculate the sample mean, we need to add up all the measurements and divide the sum by the sample size. Adding up the measurements, we get:

3.4 + 2.5 + 4.8 + 2.9 + 3.6 + 2.8 + 3.3 + 5.6 + 3.7 + 2.8 + 4.4 + 4.0 + 5.2 + 3.0 + 4.8 = 56.8

Dividing this sum by the sample size of 15, we find the sample mean to be:

56.8 / 15 = 3.79

Therefore, the sample mean for these data is 3.79.

(c) To calculate the sample median, we need to arrange the measurements in ascending order. The measurements in ascending order are:

2.5, 2.8, 2.8, 2.9, 3.0, 3.3, 3.4, 3.6, 3.7, 4.0, 4.4, 4.8, 4.8, 5.2, 5.6

Since the sample size is odd (15), the median is the middle value. In this case, the median is the 8th measurement, which is 3.4.

Therefore, the sample median for these data is 3.4.

(d) A dot plot is a simple way to visualize the data. Each measurement is represented by a dot on a number line. Here's the dot plot for the given data set:

2.5 •

2.8 ••

2.9 •

3.0 •

3.3 •

3.4 •

3.6 •

3.7 •

4.0 •

4.4 •

4.8 ••

5.2 •

5.6 •

(e) To compute the 20% trimmed mean, we need to remove the highest and lowest 20% of the data and then calculate the mean of the remaining values. Since the sample size is 15, the highest and lowest 20% would be 3 measurements.

Removing the highest and lowest 20% (3 measurements) gives us the following trimmed data set:

2.8, 2.8, 2.9, 3.0, 3.3, 3.4, 3.6, 3.7, 4.0, 4.4

Calculating the mean of these trimmed measurements, we get:

(2.8 + 2.8 + 2.9 + 3.0 + 3.3 + 3.4 + 3.6 + 3.7 + 4.0 + 4.4) / 10 = 3.44

Therefore, the 20% trimmed mean for the given data set is 3.44.

(f) The trimmed mean is less influenced by extreme values compared to the sample mean. In this case, the trimmed mean is 3.44, which is closer to the center of the remaining values after removing the extreme values. This makes the trimmed mean a more robust measure of center than the sample mean for these data.

Explanation:

answered
User Earlyadopter
by
8.3k points
1 vote

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint: 3.4 2.8 4.4 2.5 3.3 4.0 4.8 5.6 5.2 2.9 3.7 3.0 3.6 2.8 4.8 Assuming that the measurements represent a random sample from a normal population, find a 95% prediction interval for the drying time for the next trial of the paint.Solution Verified Step 1 We have the following data of drying time:\begin{align*} & 3.4, \ 2.5, \ 4.8, \ 2.9, \ 3.6, \ 2.8, \ 3.3, \ 5.6, \\ & 3.7, \ 2.8, \ 4.4, \ 4.0, \ 5.2, \ 3.0, \ 4.8 \end{align*}​3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6,3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8 hope this helps you

answered
User Isaac Gonzalez
by
8.4k points
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