Question

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QUESTION 1 An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas. They test a sample of five cars and obtain a mean of 28.2 miles. Assuming that the standard deviation is 2.7 miles, find the 95 percent confidence interval for the mean distance traveled by all such cars using 1 gallon of gas. 25.83, 30.57 24.85, 31.55 26.16, 30.24 20.70, 35.70 QUESTION 2 A federal bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years. A random sample of 20 defaulted loans yielded a mean of $67,918 with a standard deviation of $16,552.40. Calculate a 90 percent confidence interval for the mean balance of defaulted loans over the past three years. 66,487, 69,349 61,519, 74,317 39,299, 96,537 61,829, 74,007 QUESTION 3 Recently, a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred, with a sample of 319 former customers contacted. Of the 319 contacted, 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. Construct a 95 percent confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event. 0.000, .196 ©.118,.244 .240,.339 0.059, 122

Answer 1

Q.1 Solution:

Given,

$\bar x$ = 28.2   

$\sigma$ = 2.7   

n = 5   

Note that, Population standard deviation($\sigma$) is known..So we use z distribution.

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.05 $\slash$ 2 = 0.025 and 1- $\alpha$ /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

$\therefore$$Z_{\alpha/2}$ = 1.96   

The margin of error is given by

E =  $\alpha$/2 * ($\sigma$ / $\sqrt{}$ n )

= 1.96 * ( 2.7/ $\sqrt{}$ 5 )

= 2.366

Now , confidence interval for mean($\mu$) is given by:

($\bar x$ - E ) <  $\mu$ <  ($\bar x$ + E)

( 28.2 - 2.366 )   <  $\mu$ <  ( 28.2+ 2.366)

( 25.83 )   <  $\mu$ <  ( 30.57 )

Required 95% confidence interval is ( 25.83 , 30.57 )

Q.2  Solution:

Given that,

n = 20

$\bar x$ = 67918   

s = 16552.40

Note that, Population standard deviation($\sigma$) is unknown..So we use t distribution.

Our aim is to construct 90% confidence interval.   

$\therefore$ c = 0.90

$\therefore$$\alpha$ = 1- c = 1- 0.90 = 0.10

$\therefore$  $\alpha$/2 = 0.10 $\slash$ 2 = 0.05

Also, d.f = 20 - 1 = 19

$\therefore$  $t_{\alpha/2,d.f.}$  =  $t_{\alpha/2,n-1}$  =  0.05,19 = 1.729

( use t table or t calculator to find this value..)

The margin of error is given by

E =  $\alpha$/2,d.f. * ( / $\sqrt{}$ n)

= 1.729* ( 16552.40/ $\sqrt{}$ 20 )

= 6399.917

Now , confidence interval for mean($\mu$) is given by:

($\bar x$ - E ) <  $\mu$ <  ($\bar x$ + E)

( 67918 - 6399.917 )   <  $\mu$ <  ( 67918 + 6399.917 )

61,519 <  $\mu$ < 74,317

Required 90% confidence interval is ( 61,519 , 74,317 )

Q.3  Solution:

Given,

n = 319 ....... Sample size

x = 29 .......no. of successes in the sample

Let $\hat p$ denotes the sample proportion.

  $\therefore$  $\hat p$ = x/n   = 29/319 = 0.091

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.025 and 1- $\alpha$ /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

$\therefore$$Z_{\alpha/2}$ = 1.96

Now , the margin of error is given by

E = $\alpha$/2 *  $\sqrt{\hat p(1-\hat p)/n}$

= 1.96 * $\sqrt{}$ [ 0.091*(1 - 0.091)/319]

= 0.032

Now the confidence interval is given by

($\hat p$ - E)  
Zp<
($\hat p$ + E)

( 0.091 - 0.032 )   ( 0.091 + 0.032 )

Zp<

0.059   0.122  

Zp<

Required 95% Confidence Interval is ( 0.059 , 0.122 )