Question

7.A random sample of 49 observations is drawn from a population with a normal distribution. If the sample mean is 265 and the sample standard deviation is 57, find the 95% confidence interval for the population mean.

8. A random sample of 49 observations is drawn from a population with a normal distribution with a standard deviation of 57. If the sample mean is 265, find the 95% confidence interval for the population mean. Compare this confidence interval with the one found in question 7. Finally, find the 99% confidence interval for the population mean. How does this compare to the 95% confidence interval?

Answer 1

Solution:

7)

  Given that,

= 265 ....... Sample mean

s = 57 ........Sample standard deviation

n = 49   ....... Sample size

Note that, Population standard deviation() is unknown..So we use t distribution. Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025

Also, n = 49

d.f= n-1 = 48

     =    =    = 2.011

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

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

  

265 - 2.011*(57/ 49)       265 + 2.011*(57/ 49)

265 - 16.3753 < < 265 + 16.3753

  248.6247  < < 281.3753

Required interval is (248.6247 , 281.3753)

8)

Given,

= 265   ....... Sample mean

   = 57 ........Population standard deviation

n = 49   ....... Sample size

Note that, Population standard deviation() is known..So we use z distribution. Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

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

= 1.96 for the 95% Confidence interval

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

  

265 - 1.96*(57/ 49)      265 + 1.96*(57/ 49)

265 - 15.96 < < 265 + 15.96

  249.04   < < 280.96

The required interval is (249.04,280.96)

Compare this confidence interval with the one found in question 7 i.e with (248.6247 , 281.3753)

Interval using t distribution is wider than interval using z distribution.

Now we find 99% confidence interval.

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.01 2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576 for the 99% Confidence interval

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

  

  

265 - 2.576*(57/ 49)      265 + 2.576*(57/ 49)

265 - 20.976 < < 265 + 20.976

  244.024 < < 285.976

The required 99% interval is (244.024,285.976)

Compare this with 95% interval (249.04,280.96)

99% interval is wider than 95% interval.

As the level of confidence increases , confidence interval becomes wider.