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Answer :
The mean and standard deviation of the given set of data are respectively; 98.26 and 0.4848
What is the sample mean and Standard deviation?
A) We are given the temperatures of the 25 females as;
97.8, 97.2, 97.4, 97.6, 97.8, 97.9, 98.0, 98.0, 97.9, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4, 98.5, 98.6, 98.6, 98.7, 98.8, 98.8, 98.9, 98.9, 99.0.
Sample mean is gotten from the formula;
x' = ∑x/n
Where n = 25
∑x = 2456.5
x' = 2456.5/25
x' = 98.26
Now, the standard deviation is gotten from the formula;
s = √([∑(x - x')²]/(n - 1))
From online calculator plugging in all the given values, we have;
s = 0.4848
B) To test the hypothesis, let us first define the hypotheses;
We are told that Critical Appraisal of 98.6°F is the Upper Limit of the Normal Body temperature. Thus, hypotheses is;
Null Hypothesis; Hâ‚€: μ = 98.6
Alternative Hypothesis; Hâ‚: μ ≠ 98.6
test statistic is gotten from;
t = (x' - μ)/(s/√n)
t = (98.26 - 98.6)/(0.4848/√25)
t = -3.51
From online p-value from t-score calculator, using two tailed hypothesis, we have; p-value = 0.0018
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Rewritten by : Jeany
Answer:
a) [tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}=98.26[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}=0.4848[/tex]
b) [tex]t=\frac{98.26-98.6}{\frac{0.4848}{\sqrt{25}}}=-3.507[/tex]
[tex]p_v =2*P(t_{24}<-3.507)=0.0018[/tex]
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 98.6 at 5% of significance.
Step-by-step explanation:
Previous concepts and data given
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
(a) Calculate the sample mean and standard deviation.
The data given is:
97.8, 97.2, 97.4, 97.6, 97.8, 97.9, 98.0, 98.0, 97.9, 98.1, 98.2, 98.3, 98.3, 98.4, 98.4, 98.4, 98.5, 98.6, 98.6, 98.7, 98.8, 98.8, 98.9, 98.9, 99.0
In order to calculate the sample mean and the sample deviation we can use the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}=98.26[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}=0.4848[/tex]
[tex]\bar X=98.26[/tex] represent the sample mean
[tex]s=0.4848[/tex] represent the sample standard deviation
n=25 represent the sample selected
[tex]\alpha=0.05[/tex] significance level
b) Test the hypothesis Upper H0:mu= 98.6 versus H1 mu not-equals 98.6
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is 98.6, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 98.6[/tex]
Alternative hypothesis:[tex]\mu \neq 98.6[/tex]
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{98.26-98.6}{\frac{0.4848}{\sqrt{25}}}=-3.507[/tex]
P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=25-1=24[/tex]
Since is a two sided test the p value would be:
[tex]p_v =2*P(t_{24}<-3.507)=0.0018[/tex]
Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the population mean is significant different from 98.6 at 5% of significance.
