Question

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1. Below are the graphs of a standard Normal distribution and a t-distribution with 3 degrees of freedom. (a) Indicated which graph is which and explain how you know. Dotted graph Solid graph- 3-2 1 (b) On the same figure sketch a graph of a t-distribution with 1 degree of freedom. 2. Find the critical value for each of the following confidence intervals: (a) 95% confidence interval with 8 degrees (b) 80% confidence interval when n-20 of freedom You want to estimate the mean fuel efficiency of Ford Focus automobiles with 99% confidence and a margin of error of no more than 1 mile per gallon. Preliminary data suggests that σ-2.4 miles per gallon is a reasonable estimate of the standard deviation for all cars of this make and model. How large a sample do you need? 3.

Answer 1

Solution :-

Question-2 :-

Here we need to calculate the critical t* value for following intervals.

(a) :-

Given , confidence interval = 80%

degrees of freedom ,df = 8

Here $\alpha =1-95%$%

$\alpha =1-0.95$

$\alpha =0.05$

Then area, $\alpha$/2 =  0.05 / 2

  $\alpha$/2 = 0.025

We can use the t-table to find the value corresponding to an area of $\alpha$=0.025 with df=8.

From t-table, t* = 2.306

Therefore, critical t* value for 95% confidence interval is t* = 2.360

(b) :-

Given , confidence interval = 95%

Number of samples, n = 20

Here  $\alpha =1-80%$%

$\alpha =1-0.80$

$\alpha =0.2$

Then area, $\alpha$/2 =0.2 / 2

  $\alpha$/2 =0.10

Now,  degrees of freedom ,df = n - 1

df = 20 - 1

df = 19

We can use the t-table to find the value corresponding to an area of $\alpha$ = 0.10 with df = 19

From t-table, t* = 1.328

Therefore, critical t* value for 95% confidence interval is t* = 1.328

Question-3 :-

Given data:

confidence interval = 99%

$\sigma =2.4$

Margin of error  $\leq 1$

Here we need to calculate the sample size.

From given condition, i.e  

Margin of error  $\leq 1$  

$Z_{\alpha /2}*\frac{\sigma }{\sqrt{n}}\leq 1$ -------------------- (1)

Here, $\alpha =1-99$%

$\alpha =1-0.99$

$\alpha =0.01$

Then,  $Z_{\alpha /2}=Z_{0.01 /2}$

$Z_{\alpha /2}=Z_{0.005}$

$Z_{\alpha /2}=2.58$

Then, from equation (1)

$Z_{\alpha /2}*\frac{\sigma }{\sqrt{n}}\leq 1$

$2.58*\frac{2.4}{\sqrt{n}}\leq 1$

$2.58*2.4\leq \sqrt{n}$

$6.192\leq \sqrt{n}$

Now consider, Square root on both sides, we get

$6.192^2 \leq n$

$n\geq 38.340$

$\therefore n= 39$

Therefore, Sample size, n = 39

Note : -

As HOMEWORKLIB RULES rule two questions are enough to answer. that's why i didn't answer question-1. if you want that also please upload that as another question. please don't give thumbs down.

Thank you.