Do the results support or contradict the common belief that the mean body temperature is 98.6degreesF? A. The results contradic
t the belief that the mean body temperature is 98.6degreesF because both the mean and the median are less than 98.6degreesF. B. The results are inconclusive because the median body temperature is not equal to 98.6degreesF while the mean is approximately equal to 98.6degreesF. C. The results are inconclusive because the mean body temperature is not equal to 98.6degreesF while the median is approximately equal to 98.6degreesF. D. The results support the belief that the mean body temperature is 98.6degreesF because both the mean and the median are equal to 98.6degreesF
1 answer:
Answer:
A. The results contradict the belief that the mean body temperature is 98.6 °F because both the mean and the median are less than 98.6 °F.
Step-by-step explanation:
Assume the temperature data were those in the table below.
You would have calculated that
Mean = 97.9 °F
Median = 97.9 °F
These observations contradict the belief that the mean body temperature is 98.6 °F.
A 2017 study found the mean oral temperature of more than 35 000 British patients was 97.9 °F.
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Using the graph, it is found that 976 passengers had carry-on luggage that weighed less than 20 lb.
<h3>Graph:</h3>The graph is not given in this problem, but an internet search indicates that the information it contains is as follows:
- 120 passengers carry luggage of 4 lb or less.
- 222 passengers carry luggage between 5 lb and 9 lb.
- 378 passengers carry luggage between 10 lb and 14 lb.
- 256 passengers carry luggage between 15 lb and 19 lb.
- 90 passengers carry luggage between 20 lb and 24 lb.
- 40 passengers carry luggage between 25 lb or more.
Hence, the number of passengers with luggage below 20 lb is:
976 passengers had carry-on luggage that weighed less than 20 lb.
A similar problem, also involving the use of graph, is given at brainly.com/question/25836450
Answer:
[0.7210,0.8189] = [72.10%, 81.89%]
Step-by-step explanation:
The sample size is
n = 200
the proportion is
p = 154/200 = 0.77
<em>Since both np ≥ 10 and n(1-p) ≥ 10 </em>
<em>We can approximate this discrete binomial distribution with the continuous Normal distribution. As the sample size is large enough, not applying the continuity correction factor makes no significant difference. </em>
The approximation would be to a Normal curve with this parameters:
<em>Mean </em>
p = 0.77
<em>Standard deviation </em>
The 90% confidence interval for the proportion would be then
where is the 10% critical value for the Normal N(0,1) distribution , this is a value such that the area under the N(0,1) curve outside the interval
is 10%=0.1
We can either use a table, a calculator or a spreadsheet to get this value.
In Excel or OpenOffice Calc we use the function
<em>NORMSINV(0.95) and we get a value of 1.645 </em>
The 90% confidence interval for the proportion is then
This means there is a 90% probability that the proportion of people who watch educational television is between 72.10% and 81.89%
If the television company wanted to publicize the proportion of viewers, do you think it should use the 90% confidence interval?
Yes, I do.