Question

hi there, having trouble with stats homework

In a random sample of 22 people, the mean commute time to work was 31.5 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a t-distribution to construct a 80% confidence interval fo the population mean u. What is the margin of error of u? Interpret the results. The confidence interval for the population mean u is ( 29.4, 33.6) (Round to one decimal place as needed.) The margin of error of u is 2.1 (Round to one decimal place as needed.) Interpret the results. * A. With 80% confidence, it can be said that the population mean commute time is between the bounds of the confidence interval. OB. With 80% confidence, it can be said that the commute time is between the bounds of the confidence interval. XC. It can be said that 80% of people have a commute time between the bounds of the confidence interval. OD. If a large sample of people are taken approximately 80% of them will have commute times between the bounds of the confidence interval.

Answer 1

1.

• The formula of CI-

S = (T- * tn-1, 1 T+ *tn-1, 이트 12 12

where,

T
Sample Mean=31.5

S
sample sd=7.3
sample size=22

tn-1,9

a
percentile value of "t-distribution" with n-1 degrees of freedom=1.3231
$\alpha :$ Level of significance=0.20

$CI=(31.5-\frac{7.3}{\sqrt{22}}*1.3231,31.5+\frac{7.3}{\sqrt{22}}*1.3231)$

CI = (29.44,33.56)

• Margin of error:
  
  S *tn-1,3 n
  
  $=frac{7.3}{\sqrt{22}}*1.3231$
  
  2.06
  
  
• Interpretation of results:
  A) With 80% confidence, it can be said that the population mean commute time is between the bounds of the Confidence Interval.

2. a) Sample Mean:

C ΙΣ 2 n 1=1

where, x_i : is the ith data point.
n: sample size

= 911.5

b) Sample Standard Deviation:

η S = Σα; - )2 = 303.5704 n 1 1=1

c) The formula of CI-

S = (T- * tn-1, 1 T+ *tn-1, 이트 12 12

where,

T
Sample Mean=911.5

S
sample sd=303.5704
sample size=12

tn-1,9

a
percentile value of "t-distribution" with n-1 degrees of freedom=3.1058
$\alpha :$ Level of significance=0.01
$CI=(911.5-\frac{303.5704}{\sqrt{12}}*3.10258,911.5+\frac{303.5704}{\sqrt{12}}*3.10258)$
$CI=(639.33,1183.67)$

3.

• B) Use a "t-distribution" because it is random sample, $\sigma$ is unknown, and the interest rates are Normally Distributed.

• The formula of CI-

  
  S = (T- * tn-1, 1 T+ *tn-1, 이트 12 12
  
  where,
  
  T
  Sample Mean=3.58
  
  S
  sample sd=0.31
  sample size=15
  
  tn-1,9
  
  a
  percentile value of "t-distribution" with n-1 degrees of freedom=2.1448
  $\alpha :$ Level of significance=0.05
  $CI=(3.58-\frac{0.31}{\sqrt{15}}*2.1449,3.58+\frac{0.31}{\sqrt{15}}*2.1449)$
  $CI=(3.41,3.75)$

• Interpretation of the results:
  C) with 95% confidence, it can be said that the population mean interest rate is between the bounds of confidence interval.

4.

• C) Use a Normal Distribution because the Population s.d. is known and the data is normally distributed.

  • IThe formula of CI-

    
    = (х – * Та, т+ Та) n n
    
    where,
    
    T
    Sample Mean=4.916
    $\sigma :$ Population S.d.=1.26
    sample size=25
    
    Τα
    $\frac{\alpha }{2}$ percentile value of Standard Normal distribution=1.6449
    $\alpha :$ Level of significance=0.10
    $CI=(4.916-\frac{1.26}{\sqrt{25}}*1.6449,4.916+\frac{1.26}{\sqrt{25}}*1.6449)$
    $CI=(4.50,5.33)$

  • Interpretation of the result:
    A) With 90% Confidence it can be said that the ppulation mean yards per carry is between the bounds of the CI.

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