Question

A meteorologist who sampled 4 thunderstorms found that the average speed at which they traveled across a certain state was 16 miles per hour. The standard deviation of the sample was 4.1 miles per hour. Round the final answers to at least two decimal places.
Required:
Find the 90% confidence interval of the mean. Assume the variable is normally distributed.

Answer 1

Answer:

$11.18 < \mu$

Step-by-step explanation:

From the information given:

A meteorologist who sampled 4 thunderstorms  of  the sample size n = 16

the average speed at which they traveled across a certain state was 16 miles per hour ; i.e Mean $\bar x$ = 16

The standard deviation $\sigma$ of the sample was 4.1 miles per hour

The objective is to find the 90% confidence interval of the mean.

To start with the degree of freedom df = n - 1

degree of freedom df = 4 - 1

degree of freedom df = 3

At 90 % Confidence interval C.I ; the level of significance will be ∝ = 1 - C.I

∝ = 1 - 0.90

∝ = 0.10

∝/2 = 0.10/2

∝/2 = 0.050

From the tables;

Now the t value when ∝/2 = 0.050 is $t_{\alpha / 2 ,df}$

$t_{0.050 \ ,\ 3} = 2.353$

The Margin of Error = $t_{\alpha / 2 ,df} \times \dfrac{s}{\sqrt{n}}$

The Margin of Error = $2.353 \times \dfrac{4.1}{\sqrt{4}}$

The Margin of Error = $2.353 \times \dfrac{4.1}{2}$

The Margin of Error = $2.353 \times 2.05$

The Margin of Error = 4.82365

The Margin of Error = 4.82

Finally; Assume the variable is normally distributed, the  90% confidence interval of the mean is;

$\overline x - M.O.E < \mu < \overline x + M.O.E$

$16 -4.82 < \mu < 16 + 4.82$

$11.18 < \mu$

Answer 2

Answer:

The  90 % confidence  interval  for the mean population is (11.176  ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83

Step-by-step explanation:

Mean = x`= 16 miles per hour

standard deviation =s= 4.1 miles per hour

n= 4

$\frac{s}{\sqrt n}$  =  4.1/√4= 4.1/2= 2.05

1-α= 0.9

degrees of freedom =n-1=  df= 3

∈ ( estimator  t with 90 % and df= 3 from t - table ) 2.353

Using Students' t - test

x`±∈ * $\frac{s}{\sqrt n}$

Putting values

16 ± 2.353 * 2.05

= 16 + 4.82365

20.824  ;        11.176

The  90 % confidence  interval  for the mean population is (11.176  ; 20.824 )

Rounding to at least two decimal places would give 11.18 , 20.83