
A study of $36$ camels showed that they could walk an average rate of $4.2$km/hr, the sample standard deviation is $.64$. Find the $90\%$ confidence interval for mean of all camels?
Answer
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Hint: To solve this question we need to know the concept of confidence interval. To solve this problem the first step is to find the significance level. The second step is to find the critical value which depends upon the significance and the z- value table. The third step is to find the margin of error and hence we get the confidence interval for mean of all camels.
Complete step-by-step answer:
The question asks us to find a $90\%$ confidence interval for all camels when the average rate is given as $4.2$ in a study of $36$ camels having the standard deviation as $0.64$. Firstly we would write what all are given to us:
$90\%$ of Confidence Interval for $\mu $ using Normal distribution
Sample mean refers to the average sample rate,
Sample Mean = $\bar{x}$ = $4.2$
Since sample size is large, sample standard deviation can be taken as an approximation of the population standard deviation.
Population Standard deviation = $\sigma $ = $0.64$
Sample size = n = 36
Significance level = $\alpha $ = $1-\text{confidence level}$
$= 1-0.9$
$= 0.1$
The Critical Value = ${{z}_{\dfrac{\alpha }{2}}}$ = ${{z}_{\dfrac{0.1}{2}}}$ = ${{z}_{0.05}}$ = $1.645$
The Critical Values are $\pm {{z}_{\dfrac{\alpha }{2}}}=\pm 1.645$
Margin of Error = E =${{z}_{\dfrac{\alpha }{2}}}\times \dfrac{\sigma }{\sqrt{n}}$
$= 1.645\times \dfrac{0.64}{\sqrt{36}}$
$= 1.645\times \dfrac{0.64}{6}$
$= 0.175$
Margin of error hence found to be, E = $0.175$
Limits of $90\%$ confidence interval are given by:
Lower limit = $\text{\bar{x} - E}$
$= 4.2-0.175$
$= 4.025$
Upper limit = $\text{\bar{x} +E}$
$= 4.2+0.175$
$= 4.375$
$90\%$ confidence level is $\text{\bar{x} }\pm \text{E}$
$= 4.2\pm 0.175$
The $90\%$ confidence interval for the mean of all camels is $\left( 4.024,4.375 \right)$.
Note: To write the critical z- value, refer to the z table properly, as with different value of significance level the critical value changes. We need to change the percentage into the fraction and convert into decimal. To convert the percentage into fraction we need to divide the number by $100$ and then change it into decimal.
Complete step-by-step answer:
The question asks us to find a $90\%$ confidence interval for all camels when the average rate is given as $4.2$ in a study of $36$ camels having the standard deviation as $0.64$. Firstly we would write what all are given to us:
$90\%$ of Confidence Interval for $\mu $ using Normal distribution
Sample mean refers to the average sample rate,
Sample Mean = $\bar{x}$ = $4.2$
Since sample size is large, sample standard deviation can be taken as an approximation of the population standard deviation.
Population Standard deviation = $\sigma $ = $0.64$
Sample size = n = 36
Significance level = $\alpha $ = $1-\text{confidence level}$
$= 1-0.9$
$= 0.1$
The Critical Value = ${{z}_{\dfrac{\alpha }{2}}}$ = ${{z}_{\dfrac{0.1}{2}}}$ = ${{z}_{0.05}}$ = $1.645$
The Critical Values are $\pm {{z}_{\dfrac{\alpha }{2}}}=\pm 1.645$
Margin of Error = E =${{z}_{\dfrac{\alpha }{2}}}\times \dfrac{\sigma }{\sqrt{n}}$
$= 1.645\times \dfrac{0.64}{\sqrt{36}}$
$= 1.645\times \dfrac{0.64}{6}$
$= 0.175$
Margin of error hence found to be, E = $0.175$
Limits of $90\%$ confidence interval are given by:
Lower limit = $\text{\bar{x} - E}$
$= 4.2-0.175$
$= 4.025$
Upper limit = $\text{\bar{x} +E}$
$= 4.2+0.175$
$= 4.375$
$90\%$ confidence level is $\text{\bar{x} }\pm \text{E}$
$= 4.2\pm 0.175$
The $90\%$ confidence interval for the mean of all camels is $\left( 4.024,4.375 \right)$.
Note: To write the critical z- value, refer to the z table properly, as with different value of significance level the critical value changes. We need to change the percentage into the fraction and convert into decimal. To convert the percentage into fraction we need to divide the number by $100$ and then change it into decimal.
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