Question

The mean and standard deviation of a random sample of 7 baby orca whales were calculated as 430 pounds and 26.9 pounds, respectively. Assuming all conditions for inference are met, which of the following is a 90 percent confidence interval for the mean weight of all baby orca whales.

a. 26.9 ± 1.895 (430/√7 )
b. 26.9 ±1.943 (430/√7)
c. 430 ±1.440 (26.9/√7)
d. 430 ± 1.895 (26.9/√7)
e. 430 ± 1.943 (26.9/√7)

Answer 1

Answer:

$430-1.943\frac{26.9}{\sqrt{7}}$    

$430+1.943\frac{26.9}{\sqrt{7}}$    

And the best option would be:

e. 430 ± 1.943 (26.9/√7)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X=430$ represent the sample mean

$\mu$ population mean (variable of interest)

s=26.9 represent the sample standard deviation

n=7 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=7-1=6$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,6)".And we see that $t_{\alpha/2}=1.943$

Now we have everything in order to replace into formula (1):

$430-1.943\frac{26.9}{\sqrt{7}}$    

$430+1.943\frac{26.9}{\sqrt{7}}$    

And the best option would be:

e. 430 ± 1.943 (26.9/√7)

Answer 2

For 90 percent confidence interval for the mean weight of all baby orca whales is,

$430\pm1.943\dfrac{26.9}{\sqrt{7} }$

Thus option e is the correct option.

Given-

Mean $X$ of the random sample is 430 pounds.

Standard deviation $s$ of the sample is 26.9 pounds.

Confidence interval is 90 percent.

The degree of freedom is sample size n-1. Thus,

$D_f=7-1$

$D_f=6$

The critical value for 90 percent confidence level is,

$t_{\frac{a}{2} }=1.943$

The confidence interval of a mean can be given by,

$X\pm t_{\frac{a}{2}}\dfrac{s}{\sqrt{n} }$

Put the value in above equation we get,

$430\pm1.943\dfrac{26.9}{\sqrt{7} }$

Taking positive sign,

$430+1.943\dfrac{26.9}{\sqrt{7} }$

Taking negative sign,

$430-1.943\dfrac{26.9}{\sqrt{7} }$

Hence, For 90 percent confidence interval for the mean weight of all baby orca whales is,

$430\pm1.943\dfrac{26.9}{\sqrt{7} }$

Thus option e is the correct option.

For more about the confidence interval, follow the link below-

https://brainly.com/question/2396419